Integral Calculus of One-Dimensional Functions: Personal Tasks and Samples
II.P. Blazhievska, R. Riba Garcia
INTEGRAL CALCULUS
OF ONE-DIMENSIONAL FUNCTIONSPersonal tasks and Samples
Interactive textbook is developed for studentswho study on technical specialities2020 a r X i v : . [ m a t h . HO ] M a r rina Blazhievska , PhD in Physics and Mathematics,Senior Lecturer, Department of Mathematical Analysisand Probability Theory, NTUU “Igor Sikorsky KyivPolytechnic Institute,” Av. Peremogy, 37, 03056 Kyiv,Ukrainee-mail: [email protected] Ricard Riba Garcia , PhD in Mathematics, AssociateProfessor, Departament de Matem`atiques, UniversitatAut`onoma de Barcelona, Edifici Cc, 08193 Bellaterra(Barcelona), Spaine-mail: [email protected]
INTEGRAL CALCULUSOF ONE-DIMENSIONAL FUNCTIONSPersonal tasks and Samples
Electronic network tutorial edition
The interactive textbook is developed for English-speaking students whose studyon Mathematical Calculus is based on classic programs of Ukrainian higher educati-onal institutions. Its structure is as follows: the first part consists of 30 personaltasks with problems in one-dimensional integration (indefinite and definite integrals,geometric applications); the second part proposes the algorithmic solutions of thesample task with graphics built in Wolfram Mathematica 11.1, and 20 video-lessonsguided by authors. c (cid:13)
I.P. Blazhievska, R. Riba Garcia, 2020 ontents
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ersonal task 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ntroduction The changes in higher education during the last decade have increased the amount ofindividual work of students. As result, there appears the necessity of clear-developedguidelines and resources for learning (on-line learning) of fundamental sciences.This textbook is created for English-speaking students whose study in Math-ematical Calculus is based on modern programs of Ukrainian higher educationalinstitutions. It includes personal tasks and samples on “Integral Calculus of One-Dimensional Functions”, one of the classic parts of Calculus course of the first yeardegree on technical specialities. All problems are the new ones; they were gener-ated/tested by applying Wolfram Mathematica technologies. Guided by authors, 20video-lessons are available for on-line learning on our educational YouTube channel.Let us be more specific. The textbook contains 30 personal tasks and a solvedsample task with guidelines (both written and interactive versions). The mathemat-ical skeleton of each personal task consists of 4 parts: • methods of the indefinite integration; • evaluation of definite integrals; • geometric applications of definite integrals, which are related to finding the met-ric characteristics of the curves, regions and solids of revolution; • integration of improper integrals.Authors illustrate on-line how to solve 20 problems from the sample task (an accessfrom the textbook employs links and QR-codes).In more detail, each personal task consists of 40 problems divided on 12 categoriesreferring to fixed techniques ([2, Chapter 6], [3, Chapters 7-8], [4, Chapters 8-10], [5,Chapter 4] and [6, Part 3]): to be solved, the proposed 4 problems use the reduction to the table of integra-tion combined with general properties of indefinite integrals, and the method ofsubstitution under the differential (see, [2, p. 267-273], [3, p. 193-202] and [6,Chapter 13]); to be solved, the proposed 4 problems use the integration of fractions withquadratic functions (see, [2, p. 286-288] and [3, p. 214-216]); to be solved, the proposed 6 problems use the integration by parts (3 commonclasses) or suitable substitutions (see, [2, p. 272-276] and [3, p. 200-203]); to be solved, the proposed 3 problems use the integration techniques for poly-nomial fractions (see, [2, p. 276-281] and [3, p. 203-211]);5 . to be solved, the proposed 3 problems use the integration of trigonometric func-tions: products and rational functions of sines and cosines with equal arguments,products of mentioned functions with different arguments (see, [2, p.281-286],[3, p. 212-214] and [6, Chapter 15]); to be solved, the proposed 2 problems use the integration techniques for fractionswith radicals by applying the trigonometric and hyperbolic substitutions (see,[2, p. 286-289], [3, p. 214-219] and [6, Chapters 15, 17]); to be solved, the proposed 4 problems use the integration techniques for definiteRiemann integrals with applying the Newton-Leibnitz formula. Here, the gen-eral evaluation, integration by parts and applying of suitable substitutions arecovered (see, [2, p. 290-300] and [3, p. 221-233]); to be solved, the proposed 3 problems require the building of correct regions inCartesian, Parametric and Polar coordinates, applying of suitable formulas andvalid evaluation of definite integrals (see, [2, p. 306-311], [3, p. 237-242] and [6,p. 233-257]); to be solved, the proposed 3 problems require the building of correct arc segmentsin Cartesian, Parametric and Polar coordinates, applying of suitable formulasand valid evaluation of definite integrals (see, [2, p. 311-314], [3, p. 242-245]and [6, p. 257-261]); to be solved, the proposed 3 problems require the building of correct generatricesin Cartesian, Parametric and Polar coordinates, analysis of surfaces of revolution,applying of suitable formulas and valid evaluation of definite integrals (see, [2,p. 314-316], [3, p. 247-248] and [6, p. 267]); to be solved, the proposed 3 problems require the building of correct rotatingregions in Cartesian and Polar coordinates, analysis of solids of revolution, ap-plying of suitable formulas and valid evaluation of definite integrals (see, [2, p.317-319], [3, p. 245-247] and [6, p. 265-267]); to be solved, the proposed 2 problems use the integration techniques for improperRiemann integrals of the 1st and 2nd kinds (see, [2, p. 300-305] and [3, p. 233-237]).The following classic list of the curves are employed in the textbook: • in Cartesian coordinates: ellipses, exponentials, circles, catenaries, logarithmicfunctions, squared and cubic parabolas (power functions), straight lines andgeneral curves; 6 in Parametric coordinates: astroids, ellipses, circles, cycloids, involutes of classiccurves and spirals; • in Polar coordinates: Archimedean and logarithmic spirals, cardioids, lemnis-cates (8-shaped and ∞ -shaped) and petaled roses.If a parametric range is omitted in the problem, it is supposed to be the whole regionof the curve’s well-definiteness.In order to visualize the shapes of the curves, regions, surfaces and solids, westrongly recommend students to use the following links: • • http://mathworld.wolfram.com; • http://old.nationalcurvebank.org/volrev/volrev.htmThe suitable formulas for evaluation of metric characteristics are available as well.The sample task consists on algorithmic solutions of the proposed 40 problems.Here, we point out the applied techniques for indefinite and definite integrals (cat-egories 1–7), develop step-by-step algorithms with 2D/3D-simulations via WolframMathematica 11.1 for geometric problems (categories 8-11) and solve the problemswith improper integrals (category 12). Based on problems from the sample task,we created 20 illustrative video-lessons on techniques of integration. These lessonswere prepared with KPI TV c (cid:13) Team and nowadays are available on our educationalYouTube channel: • n notions used in the textbook Following by mathematical traditions stated in Ukraine (Post-Soviet countries), wehave used the following notions for some trigonometric, hyperbolic functions andtheir inverses: • tg x instead of tan x, tg x = sin x cos x ; • ctg x instead of cot x, ctg x = cos x sin x ; • arctg x instead of arctan x ; • arcctg x instead of arccot x, arcctg x = π − arctg x ; • sh x instead of sinh x, sh x = e x − e − x • ch x instead of cosh x, ch x = e x + e − x • th x instead of tanh x, th x = sh x ch x ; • cth x instead of coth x, cth x = ch x sh x ; • arcsh x instead of arcsinh x, arcsh x = ln (cid:0) x + √ x + 1 (cid:1) ; • arcch x instead of arccosh x, arcch x = ln (cid:0) x + √ x − (cid:1) , | x | ≥ • arcth x instead of arctanh x, arcth x = 12 ln (cid:18) x − x (cid:19) , | x | < • arccth x instead of arccoth x, arccth x = 12 ln (cid:18) x + 1 x − (cid:19) , | x | > For more information about hyperbolic functions, inverse hyperbolic functionsand their transforms see [2, p. 285-289] and [6, Chapter 17].8 ersonal task 1 Integrate using the table andsubstitution under differential: (cid:82) x (3 x − dx ;a) (cid:82) tg xdx ;b) (cid:82) sin xe x − dx ;c) (cid:82) cos (1 − x ) dx .d) Integrate the quadratic fractions: (cid:82) − xx − dx ;a) (cid:82) dx √
20 + 24 x − x ;b) (cid:82) ( x + 4) dxx + 6 x + 5 c) (cid:82) (2 x − dx √ x − x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) dx √ e x + 1 ;a) (cid:82) dxx √ x − ;b) (cid:82) ln( x − dx ;c) (cid:82) ( x + 5) cos 2 xdx ;d) (cid:82) arcsin 2 xdx ;e) (cid:82) ln(cos x )cos x dx .f) Integrate the polynomial fractions: (cid:82) x + 14 x + 19( x + 4 x + 3)( x + 5) dx ;a) (cid:82) x + 1 x − x + x dx ;b) (cid:82) x + 6( x + 2 x + 5)( x − dx .c) Integrate trigonometric expressions: (cid:82) sin 3 x cos xdx ;a) (cid:82) cos x sin xdx ;b) (cid:82) dx x + 3 cos x .c) Integrate the fractions with radicals: (cid:82) √ − x x dx ;a) (cid:82) (1 − √ x + 1) dx (1 + √ x + 1) √ x + 1 .b) Solve the definite integrals: e +1 (cid:82) x ln( x − dx ;a) π/ (cid:82) (cid:112) tg x cos x dx ;b) √ (cid:82) x √ x dx ;c) (cid:82) (cid:112) ( x − dx (cid:112) ( x − . d) Find the area of the figure boundedby the curves: y = 2 x − x + 6 , y = x − x ;a) (cid:26) x = 4 cos t,y = 4 sin t ; x = 2 ( x ≥ ;b) ρ = 6 cos 3 φ. c) Find the arc-length of the curve: y = ln x, √ ≤ x ≤ √ ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = x , − ≤ x ≤ , l = OX ;a) (cid:26) x = 1 + 2 cos t,y = 3 + 2 sin t ; ( x ≥ l = OY ;b) ρ = 2 (cid:113) sin 2 (cid:0) φ − π (cid:1) , l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 1 , y = x + 1 , l = OX ;a) x = − y + 5 y − , x = 0 , l = OY ;b) ρ = 2(1 + cos φ ) , l = oρ .c) Solve the improper integrals: ∞ (cid:82) xx + 16 dx ;a) / (cid:82) dx √ − x .b)9 ersonal task 2 Integrate using the table andsubstitution under differential: (cid:82) (3 − √ x ) dx √ x ;a) (cid:82) (cos 2 x ) dx cos x sin x ;b) (cid:82) x x − dx ;c) (cid:82) sin (1 − x ) dx .d) Integrate the quadratic fractions: (cid:82) x + 5 x − dx ;a) (cid:82) dx √ − x − x ;b) (cid:82) (2 x − dxx + 6 x + 13 ;c) (cid:82) (4 x + 1) dx √ x + 4 x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) (4 x + 3) dx ( x − ;a) (cid:82) dxx √ x + 4 x + 1 ;b) (cid:82) (1 − x + x ) sin 3 xdx ;c) (cid:82) x ln xdx ;d) (cid:82) arctg x dx ;e) (cid:82) cos(ln x ) dx .f) Integrate the polynomial fractions: (cid:82) x + 15 x − x + x − x − dx ;a) (cid:82) x − x − x + 1 x − x dx ;b) (cid:82) x + 4 x − x − x + 8)( x − dx .c) Integrate trigonometric expressions: (cid:82) sin 3 x sin 9 xdx ;a) (cid:82) cos x √ sin xdx ;b) (cid:82) dx x − x .c) Integrate the functions with radicals: (cid:82) x √ − x dx ;a) (cid:82) (cid:114) x + 4 x − dx ( x + 4) .b) Solve the definite integrals: (cid:82) ( x − e x dx ;a) π (cid:82) cos ( x ) sin ( x ) dx ;b) √ e − (cid:82) x x + 1 dx ;c) ln 2 (cid:82) dxe x (3 + e − x ) . d) Find the area of the figure boundedby the curves: y = ( x − , y = 4 x − ;a) (cid:26) x = 3 cos t,y = 4 sin t ; y = 2 ( y ≥ ;b) ρ = sin 2 φ. c) Find the arc-length of the curve: y = √ − x , − √ ≤ x ≤ ;a) (cid:26) x = 4 e t cos t,y = 4 e t sin t ; 0 ≤ t ≤ π ;b) ρ = 2(1 − cos φ ) .c) Find the area of the surface formedby rotating the curves around the l -axis: y = e x , ≤ x ≤ ln 8 , l = OX ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π,l = OX ; b) ρ = 3 √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x + 2 x + 1 , y = 1 − x, l = OX ;a) x = − y + 2 y, x = 4 y − y , l = OY ;b) ρ = 4 sin φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x x − dx ;a) π/ (cid:82) π/ sin x cos x dx .b)10 ersonal task 3 Integrate using the table andsubstitution under differential: (cid:82) √ x (1 − x ) dx ;a) (cid:82) ctg xdx ;b) (cid:82) ln (2 x − x − dx ;c) (cid:82) cos ( x + 3) dx .d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) dx √−
55 + 16 x − x ;b) (cid:82) (3 x + 4) dxx + 2 x + 5 ;c) (cid:82) (2 x + 3) dx √ x + 2 x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) e x √ e x + 1 dx ;a) (cid:82) √ x x dx ;b) (cid:82) x sin 2 xdx ;c) (cid:82) x + 6 x − e x dx ;d) (cid:82) arccos 4 xdx ;e) (cid:82) x cos x dx .f) Integrate the polynomial fractions: (cid:82) x − x ( x − x + 3)( x − dx ;a) (cid:82) x + 3 x − x + 12 x + 4 x dx ;b) (cid:82) − x + 24 x − x − x + 13)( x + 1) dx .c) Integrate trigonometric expressions: (cid:82) cos 7 x cos 3 xdx ;a) (cid:82) sin x cos x dx ;b) (cid:82) cos xdx (sin x + 1) − cos x .c) Integrate the fractions with radicals: (cid:82) dxx √ x − ;a) (cid:82) (cid:112) ( x + 1) + √ x + 1 √ x + 1 + (cid:112) ( x + 1) dx .b) Solve the definite integrals: e − (cid:82) ln ( x + 1) x + 1 dx ;a) π (cid:82) sin xdx ;b) √ (cid:82) x x dx ;c) (cid:82) dx x + √ x + 1 . d) Find the area of the figure boundedby the curves: x + y = 7 , xy = 6 ;a) (cid:26) x = 2 cos t,y = 4 sin t ; y = 0 ( y ≥ ;b) ρ = √ sin 2 φ. c) Find the arc-length of the curve: y = 13 x , − ≤ x ≤ ;a) (cid:26) x = 4( t − sin t ) ,y = 4(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = e φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = ch x, ≤ x ≤ ln 8 , l = OX ;a) (cid:26) x = 2 + cos t,y = sin t ; l = OY ;b) ρ = 6 cos φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 5 , y = x + 5 , l = OX ;a) y = ln x, x = 0 , y = 0 , y = ln 4 , l = OY ;b) ρ = 3(1 − cos φ ) , l = oρ .c) Solve the improper integrals: ∞ (cid:82) x (cid:112) ( x + 1) dx ;a) π/ (cid:82) cos x sin x dx .b)11 ersonal task 4 Integrate using the table andsubstitution under differential: (cid:82) √ x (1 − x ) dx ;a) (cid:82) (sin 2 x ) dx (cos x − − ;b) (cid:82) x e x − dx ;c) (cid:82) th xdx .d) Integrate the quadratic fractions: (cid:82) − x − x − dx ;a) (cid:82) dx √
12 + 12 x − x ;b) (cid:82) ( − x + 3) dxx − x + 10 ;c) (cid:82) (4 x − dx √ x + 2 x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) x + 5( x + 3) dx ;a) (cid:82) dx ( x + 1) √− x − x ;b) (cid:82) x (cos x + sin x ) dx ;c) (cid:82) ln( x + 1) x + 1 dx ;d) (cid:82) sin(ln x ) dx ;e) (cid:82) arctg x dx .f) Integrate the polynomial fractions: (cid:82) x + 42( x + 5 x + 4)( x − dx ;a) (cid:82) x − x − x + 2 x + x dx ;b) (cid:82) x + 22 x + 27( x + 4 x + 5)( x + 2) dx .c) Integrate trigonometric expressions: (cid:82) cos 4 x sin 7 xdx ;a) (cid:82) cos x sin x dx ;b) (cid:82) dx − x + 2 cos x .c) Integrate the fractions with radicals: (cid:82) √ − x x dx ;a) (cid:82) √ x + √ x √ x dx .b) Solve the definite integrals: (cid:82) / ln 5 e − x x dx ;a) π/ (cid:82) cos x sin xdx ;b) (cid:82) x + 5 x − dx ;c) (cid:82) − x √ − x dx. d) Find the area of the figure boundedby the curves: y = ( x + 3) , y = 2 x + 6 ;a) (cid:26) x = 4 cos t,y = 4 sin t ; x = 2 ( x ≥ ;b) ρ = sin 3 φ. c) Find the arc-length of the curve: y = ln(sin x ) , π ≤ x ≤ π ;a) (cid:26) x = 2 cos t − cos 2 t,y = 2 sin t − sin 2 t ; 0 ≤ t ≤ π ;b) ρ = φ, ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = √ x + 6 , ≤ x ≤ , l = OX ;a) (cid:26) x = cos t,y = sin t ; l = OY ;b) ρ = 3 √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 4 , y = x, l = OX ;a) x = 9 − ( y − , x = 0 , l = OY ;b) ρ = 6 sin φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x − x − x − dx ;a) (cid:82) dx √ − x − x .b)12 ersonal task 5 Integrate using the table andsubstitution under differential: (cid:82) √ x (3 + 2 x √ x ) dx ;a) (cid:82) dx cos 2 x + sin x ;b) (cid:82) x dx √ x + 9 ;c) (cid:82) (1 − sin x ) dx .d) Integrate the quadratic fractions: (cid:82) − x + 7 x − dx ;a) (cid:82) dx √
15 + 6 x − x ;b) (cid:82) ( − x − dxx − x + 20 ;c) (cid:82) (6 x − dx √ x + 4 x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) dx √ e x ;a) (cid:82) dxx √ x + 4 ;b) (cid:82) ln xdx ;c) (cid:82) (1 − x ) cos 3 xdx ;d) (cid:82) e √ x +3 dx ;e) (cid:82) x · arctg xdx .f) Integrate the polynomial fractions: (cid:82) − x − x + 9( x + 4 x + 3)( x + 2) dx ;a) (cid:82) x + 3 x + 3 x + 4 x + 4 x + 4 x dx ;b) (cid:82) x − x + 24( x − x + 10)( x − dx .c) Integrate trigonometric expressions: (cid:82) sin 2 x sin 4 xdx ;a) (cid:82) (cid:114) cos x sin x dx ;b) (cid:82) cos xdx − sin x + 3 cos x .c) Integrate the functions with radicals: (cid:82) x √ − x dx ;a) (cid:82) (cid:114) x + 1 x − dx ( x − .b) Solve the definite integrals: (cid:82) (2 − x ) e x dx ;a) π/ (cid:82) − π/ sin x cos x dx ;b) (cid:82) − / dx √ x − x ;c) (cid:82) (cid:112) ( x + 1) dx (cid:112) ( x + 1) . d) Find the area of the figure boundedby the curves: x + y = − , xy = 2 ;a) (cid:26) x = 5 cos t,y = 2 sin t ; y = 0 ( y ≥ ;b) ρ = 4(1 + sin φ ) .c) Find the arc-length of the curve: y = 3 + ch x, ≤ x ≤ ln 2 ;a) (cid:26) x = 4 cos t,y = 4 sin t ; 0 ≤ t ≤ π ;b) ρ = √ e φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = x , ≤ x ≤ √ , l = OY ;a) (cid:26) x = 3( t − sin t ) ,y = 3(1 − cos t ); t ≤ π,l = OX ;b) ρ = 8 sin φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 2 , y = x + 2 , l = OX ;a) x = y + 4 , x = 0 , | y | = 2 , l = OY ;b) ρ = 2 sin 2 φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) / √ x x + 1 dx ;a) e (cid:82) dxx (cid:112) − ln x .b)13 ersonal task 6 Integrate using the table andsubstitution under differential: (cid:82) x (6 √ x − √ x ) dx a) (cid:82) (sin 2 x ) dx (2 − sin x ) − ;b) (cid:82) ln (3 x + 1)3 x + 1 dx ;c) (cid:82) cos (2 x − dx .d) Integrate the quadratic fractions: (cid:82) x + 7 x − dx ;a) (cid:82) dx √ x − x + 24 ;b) (cid:82) ( − x + 1) dxx + 6 x + 13 ;c) (cid:82) − x + 3 √− x − x dx .d) Integrate by parts or using thesuitable substitutions: (cid:82) e x dx √ e x + 16 ;a) (cid:82) dxx √ − x + x ;b) (cid:82) (3 − x ) sin 4 xdx ;c) (cid:82) x − e x dx ;d) (cid:82) (arcsin x ) dx ;e) (cid:82) ln xx dx .f) Integrate the polynomial fractions: (cid:82) x + 14( x − x + 4)( x − dx ;a) (cid:82) x + 7 x + 9 x + 9 x + 3 x dx ;b) (cid:82) x − x + 46( x − x + 10)( x + 2) dx .c) Integrate trigonometric fractions: (cid:82) sin 7 x cos 2 xdx a) (cid:82) sin x cos x dx ;b) (cid:82) xdx −
12 cos x − x .c) Integrate the fractions with radicals: (cid:82) √ x − x dx ;a) (cid:82) (cid:114) x + 2 x − dx ( x + 2) .b) Solve the definite integrals: e +2 (cid:82) ln ( x − x − dx ;a) π/ (cid:82) − π/ cos x sin xdx ;b) (cid:82) x − x dx ;c) (cid:82) − √ x + 44 + √ x + 4 dx. d) Find the area of the figure boundedby the curves: x + 2 y = 5 , xy = 2 ;a) (cid:26) x = 3 cos t,y = 3 sin t ; x = 0 ( x ≥ ;b) ρ = 2 cos 4 φ. c) Find the arc-length of the curve: y = ln( x + √ x − , ≤ x ≤ √ ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = 6 cos φ − φ .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 1 + ch x, − ≤ x ≤ , l = OX ;a) (cid:26) x = 1 + 2 cos t,y = 2 sin t ; ( x ≥ , l = OY ;b) ρ = 1 + cos φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 3 , y = x + 3 , l = OX ;a) y = √ − x , y = x, x = 0 , l = OY ;b) ρ = 4 φ, ≤ φ ≤ π, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x x + 1 dx ;a) π/ (cid:82) e − ctg x dx sin x .b)14 ersonal task 7 Integrate using the table andsubstitution under differential: (cid:82) ( x − √ x dx ;a) (cid:82) tg x cos x dx ;b) (cid:82) x √ x − dx ;c) (cid:82) (cid:16) x (cid:17) dx ;d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) dx √ − x − x ;b) (cid:82) ( − x + 9) dxx + 4 x + 20 ;c) (cid:82) (6 x + 3) dx √ x − x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) dxx √ x + 4 ;a) (cid:82) √ xx ln x dx ;b) (cid:82) (2 − x ) sin 2 xdx ;c) (cid:82) x x dx ;d) (cid:82) arcctg √ xdx ;e) (cid:82) ln( x + 1) dx .f) Integrate the polynomial fractions: (cid:82) x + 5 x − x + 4 x + 3)( x + 2) dx ;a) (cid:82) x + 11 x + 17 x + 9 x + 5 x + 8 x + 4 dx ;b) (cid:82) x + x − x + 2 x + 2 x dx .c) Integrate trigonometric expressions: (cid:82) cos x sin x cos 3 xdx ;a) (cid:82) cos x √ sin xdx ;b) (cid:82) dx − x .c) Integrate the functions with radicals: (cid:82) √ − x − x dx ;a) (cid:82) dxx ( √ x + √ x ) .b) Solve the definite integrals: ln 2 (cid:82) − ( x + 3) e x dx ;a) π (cid:82) cos ( x ) sin ( x ) dx ;b) (cid:82) x x + 1 dx ;c) − ln 2 (cid:82) √ − e x dx. d) Find the area of the figure boundedby the curves: y = ( x + 1) , y = 6 x + 6 ;a) (cid:26) x = 12 cos t,y = 5 sin t ; x = 6 ( x ≥ ;b) ρ = 2 √ cos 2 φ. c) Find the arc-length of the curve: y = ln(cos 2 x ) , ≤ x ≤ π ;a) (cid:26) x = 4 cos t,y = 4 sin t ; y = 2 ( y ≥ ;b) ρ = 6 sin (cid:16) φ (cid:17) .c) Find the area of the surface formedby rotating the curves around the l -axis: y = √ x + 7 , ≤ x ≤ √ , l = OX ;a) (cid:26) x = 3 cos t,y = 3 sin t ; l = OY ;b) ρ = 2 e − φ , ≤ φ ≤ π, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = ( x − , y = x + 4 , l = OX ;a) y = 9 − x , y = 9 − x , y = 0 , l = OY ;b) ρ = 4 sin 2 φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) dxx ( x + 1) ;a) /π (cid:82) sin (cid:18) x (cid:19) dxx .b)15 ersonal task 8 Integrate using the table andsubstitution under differential: (cid:82) x − x e x + 4 x dx ;a) (cid:82) x x dx ;b) (cid:82) ln (3 − x )3 − x dx ;c) (cid:82) (1 − sin 3 x ) dx .d) Integrate the quadratic fractions: (cid:82) − x + 3 x − dx ;a) (cid:82) dx √ x − x + 8 ;b) (cid:82) ( − x + 1) dxx − x + 17 ;c) (cid:82) (2 x + 3) dx √ − x − x .d) Integrate by parts or using thesuitable substitutions: (cid:82) ln xx (cid:112) − ln x dx ;a) (cid:82) √ x + 1 x dx ;b) (cid:82) (3 − x ) sin 3 xdx ;c) (cid:82) (1 − x ) e x dx ;d) (cid:82) arctg xx dx ;e) (cid:82) x sin x dx .f) Integrate the polynomial fractions: (cid:82) x − x − x + 6 x + 5)( x − dx ;a) (cid:82) x + 3 x + 4 x − x + 5 x + 8 x + 4 dx ;b) (cid:82) − x − x − x + 4 x + 5)( x − dx .c) Integrate trigonometric expressions: (cid:82) sin 7 x sin 4 xdx ;a) (cid:82) cos x sin xdx ;b) (cid:82) − cos x x dx .c) Integrate the fractions with radicals: (cid:82) dxx √ x − x ;a) (cid:82) xdx ( √ x + 2 √ x ) .b) Solve the definite integrals: e − (cid:82) ( x + 1) ln( x + 1) dx ;a) π/ (cid:82) cos x sin xdx ;b) (cid:82) x − √ x + 1 dx ;c) π/ (cid:82) e tg x dx cos x . d) Find the area of the figure boundedby the curves: y = ( x − , y = 4( x − ;a) (cid:26) x = 3 cos t,y = sin t ; y = 0 ( y ≥ ;b) ρ = 2 sin 3 φ. c) Find the arc-length of the curve: y = 4 − ch x, ≤ x ≤ ln 4 ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = √ e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 2 x , ≤ x ≤ √ , l = OY ;a) (cid:26) x = 1 + 3 cos t,y = 3 sin t ; l = OX ;b) ρ = 4 √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 4 x, y = 0 , l = OX ;a) y = x , y = 5 x − , l = OY ;b) ρ = 2(1 − cos φ ) , l = oρ .c) Solve the improper integrals: ∞ (cid:82) − arcctg x x dx ;a) √ (cid:82) x + 2 √ x + 2 x − dx .b)16 ersonal task 9 Integrate using the table andsubstitution under differential: (cid:82) x (1 − √ x ) dx ;a) (cid:82) tg x − cos x dx ;b) (cid:82) arcsin x + x √ − x dx ;c) (cid:82) (sin x + cos x ) dx. d) Integrate the quadratic fractions: (cid:82) − x + 4 x − dx ;a) (cid:82) dx √ − x − x ;b) (cid:82) ( − x − dxx + 6 x + 13 ;c) (cid:82) (6 x + 8) dx √ x + 4 x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) x − x − dx ;a) (cid:82) dx ( x − √ x − x ;b) (cid:82) x (sin x − x ) dx ;c) (cid:82) x − x + 1 e x dx ;d) (cid:82) ln( x − x − dx ;e) (cid:82) ( arccos x ) dx .f) Integrate the polynomial fractions: (cid:82) x + 14 x − x − x + 5)( x + 3) dx ;a) (cid:82) x + 15 x + 24 x + 10 x + 5 x + 7 x + 3 dx ;b) (cid:82) x − x + 30( x − x + 20)( x + 2) dx .c) Integrate trigonometric expressions: (cid:82) cos x sin 8 xdx ;a) (cid:82) cos x sin x dx ;b) (cid:82) dx x + 5 cos x .c) Integrate the fractions with radicals: (cid:82) √ − x x dx ;a) (cid:82) (cid:114) − x x dxx .b) Solve the definite integrals: /π (cid:82) /π cos (cid:16) x (cid:17) dxx ;a) π/ (cid:82) x cos x sin xdx ;b) √ (cid:82) x − x + 9 dx ;c) (cid:82) x √ − x dx. d) Find the area of the figure boundedby the curves: y = − x + 5 x + 1 , y = 5 x ;a) (cid:26) x = 1 + 2 cos t,y = 2 sin t ; y = 1 ( y ≥ ;b) ρ = cos 2 φ. c) Find the arc-length of the curve: y = 3 ln(9 − x ) , ≤ x ≤ ;a) (cid:26) x = 3 e t cos t,y = 3 e t sin t ; 0 ≤ t ≤ π ;b) ρ = 4 cos (cid:16) φ (cid:17) .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 3 − ch x, − ≤ x ≤ , l = OX ;a) (cid:26) x = 2 cos t,y = 2 sin t ; l = OY ;b) ρ = 6 sin φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x + 6 x + 9 , y = − x, l = OX ;a) y = x − , y = 7 x, x = 0 ( x ≥ ,l = OY ;b) ρ = 2 φ, ≤ φ ≤ π, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x e x dx ;a) (cid:82) x − x √ − x dx .b)17 ersonal task 10 Integrate using the table andsubstitution under differential: (cid:82) x √ x (1 − x √ x ) dx ;a) (cid:82) cos 2 x cos x dx ;b) (cid:82) − x √ − x dx ;c) (cid:82) (1 + cos x ) dx .d) Integrate the quadratic fractions: (cid:82) − x + 5 x − dx ;a) (cid:82) dx √ x − x − ;b) (cid:82) ( − x + 2) dxx − x + 17 ;c) (cid:82) − x + 7 √ x − x dx .d) Integrate by parts or using thesuitable substitutions: (cid:82) − e x e x dx ;a) (cid:82) dxx √ x − ;b) (cid:82) (2 − x ) cos 2 xdx ;c) (cid:82) ( x + 1) ln( x + 1) dx ;d) (cid:82) sin √ xdx ;e) (cid:82) x arctg x x dx .f) Integrate the polynomial fractions: (cid:82) x + 12 x + 6( x + 3 x + 2)( x + 3) dx ;a) (cid:82) x − x − x + 1 x − x − x + 1 dx ;b) (cid:82) − x − x + 15( x + 4 x + 5)( x − dx .c) Integrate trigonometric expressions: (cid:82) cos x sin x cos 3 xdx ;a) (cid:82) √ cos x sin xdx ;b) (cid:82) dx sin x + cos x .c) Integrate the fractions with radicals: (cid:82) x (cid:112) (1 − x ) dx ;a) (cid:82) xdx (2 + x ) √ x .b) Solve the definite integrals: ln 4 − (cid:82) − ( x + 1) e x +2 dx ;a) π (cid:82) sin x cos xdx ;b) (cid:82) − x − √ − x − x dx ;c) √ (cid:82) √ x + 1 + 1 √ x + 1 − dx. d) Find the area of the figure boundedby the curves: x + y = 5 , xy = 2 ;a) (cid:26) x = 3 cos t,y = 6 sin t ; x = 0 ( x ≥ ;b) ρ = 2(1 + sin φ ) .c) Find the arc-length of the curve: y = 5 − ch x, ≤ x ≤ ln 3 ;a) (cid:26) x = 2 cos t,y = 2 sin t ; 0 ≤ t ≤ π ;b) ρ = 5 e φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 2 x , ≤ x ≤ √ , l = OY ;a) (cid:26) x = t − sin t,y = 1 − cos t ; 0 ≤ t ≤ π, l = OX ;b) ρ = 3 (cid:113) sin 2 (cid:0) φ − π (cid:1) , l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x, y = 0 , l = OX ;a) y = 9 − x , y = 8 x, x = 0 ( x ≥ ,l = OY ;b) ρ = 2 cos φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x − x + 4 dx ;a) e (cid:82) ln x − x (cid:112) − ln x dx .b)18 ersonal task 11 Integrate using the table andsubstitution under differential: (cid:82) ( − x + 3) dxx ;a) (cid:82) ( tg x + ctg x ) dx ;b) (cid:82) x e − x +1 dx ;c) (cid:82) sin ( x + 2) dx .d) Integrate the quadratic fractions: (cid:82) − xx − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) ( x + 2) dxx + 8 x + 7 ;c) (cid:82) ( − x − dx √ x − x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) x − x √ x + 1 dx ;a) (cid:82) √ − x x dx ;b) (cid:82) x cos 4 xdx ;c) (cid:82) arcsin 3 xdx ;d) (cid:82) e √ x dx ;e) (cid:82) ln( x + √ x + 1) dx .f) Integrate the polynomial fractions: (cid:82) x + 7 x − x + 3 x + 2)( x − dx ;a) (cid:82) x + 2 x − x − x − x − x dx ;b) (cid:82) x + 14 x + 16( x + 2 x + 2)( x + 2) dx .c) Integrate trigonometric expressions: (cid:82) sin 5 x cos 2 xdx ;a) (cid:82) cos x sin xdx ;b) (cid:82) dx − x + cos x .c) Integrate the fractions with radicals: (cid:82) √ x − x dx ;a) (cid:82) dx ( √ x + 3 − √ x + 3 .b) Solve the definite integrals: e (cid:82) (2 x −
1) ln xdx ;a) π/ (cid:82) (cid:112) tg x cos x dx ;b) (cid:82) x √ x dx ;c) (cid:82) √ x + 15 + √ x + 1 dx. d) Find the area of the figure boundedby the curves: y = − x + 4 x + 1 , y = − x + 5 ;a) (cid:26) x = 2 cos t,y = 2 sin t ; x = 0 ( x ≥ ;b) ρ = 3 cos 2 φ .c) Find the arc-length of the curve: y = 4 ln x, √ ≤ x ≤ √ ;a) (cid:26) x = 3( t − sin t ) ,y = 3(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = 3 e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = x , − ≤ x ≤ , l = OX ;a) (cid:26) x = 2 + sin t,y = 2 + cos t ; l = OY ;b) ρ = 6 sin (cid:16) φ (cid:17) , l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x + 4 x + 4 , y = 3 x + 6 , l = OX ;a) x = − y + 4 y, x = 0 , l = OY ;b) ρ = 2 sin φ, l = oρ .c) Solve the improper integrals: (cid:82) −∞ x x + 4 dx ;a) / (cid:82) dx (cid:112) (1 − x ) .b)19 ersonal task 12 Integrate using the table andsubstitution under differential: (cid:82) x x − x x dx ;a) (cid:82) (5 − ctg x ) dx ;b) (cid:82) x x − dx ;c) (cid:82) cos (3 x + 1) dx ;d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) (2 x − dxx − x + 13 c) (cid:82) ( − x + 3) dx √ x + 2 x − ;d) Integrate by parts or using thesuitable substitutions: (cid:82) x ( x + 2) dx ;a) (cid:82) dxx √ x + 2 x + 1 ;b) (cid:82) (4 − x ) sin 4 xdx ;c) (cid:82) ( x − ln( x − dx ;d) (cid:82) e √ x − dx ;e) (cid:82) x arcctg xx + 1 dx .f) Integrate the polynomial fractions: (cid:82) x − x − x − x + 2)( x + 2) dx ;a) (cid:82) − x − x − x + 70( x + 2)( x + 3 x − dx ;b) (cid:82) x + 6 x + 10 x + 2 x + 2 x dx ;c) Integrate trigonometric expressions: (cid:82) sin 2 x sin 5 xdx a) (cid:82) cos x √ sin xdx ;b) (cid:82) − dx − x ;c) Integrate the fractions with radicals: (cid:82) √ − x x dx ;a) (cid:82) (cid:114) x + 1 x − dx ( x + 1) b) Solve the definite integrals: (cid:82) − x e x +2 dx ;a) π/ (cid:82) cos ( x ) sin ( x ) dx ;b) / √ (cid:82) arcsin x + x √ − x dx ;c) e − (cid:82) ln( x + 1) dx √ x + 1 . d) Find the area of the figure boundedby the curves: y = ( x + 3) , y = 4 x + 12 ;a) (cid:26) x = 3( t − sin t ) ,y = 3(1 − cos t ); y = 0 , ≤ t ≤ π ;b) ρ = 2 sin 2 φ. c) Find the arc-length of the curve: y = 2 ln(4 − x ) , − ≤ x ≤ ;a) (cid:26) x = 2 e t cos t,y = 2 e t sin t ; 0 ≤ t ≤ π ;b) ρ = 4(1 − cos φ ) .c) Find the area of the surface formedby rotating the curves around the l -axis: y = (cid:114) − x , − ≤ x ≤ , l = OX ;a) (cid:26) x = cos t,y = sin t ; l = OY ;b) ρ = √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 4 x + 1 , y = 5 − x, l = OX ;a) x = − y + 7 y, x = 6 , l = OY ;b) ρ = 6 φ, ≤ φπ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x − x + 9 dx ;a) (cid:82) arccos x √ − x dx .b)20 ersonal task 13 Integrate using the table andsubstitution under differential: (cid:82) (2 − x √ x ) dxx ;a) (cid:82) ( tg x − ctg x ) dx ;b) (cid:82) ln (2 x + 3) dx x + 3 ;c) (cid:82) cos (3 − x ) dx ;d) Integrate the quadratic fractions: (cid:82) x + 3 x − dx ;a) (cid:82) dx √
12 + 12 x − x ;b) (cid:82) (4 x + 6) dxx − x + 8 ;c) (cid:82) (2 x − dx √ x − x − ;d) Integrate by parts or using thesuitable substitutions: (cid:82) dx √ − e x ;a) (cid:82) √ x x dx ;b) (cid:82) x sin 3 xdx ;c) (cid:82) x + 4 xe x dx ;d) (cid:82) arccos 2 xdx ;e) (cid:82) x + 3cos x dx ;f) Integrate the polynomial fractions: (cid:82) x − x + 25( x − x + 6)( x − dx ;a) (cid:82) x − x + 52 x + 49 x − x + 49 x dx ;b) (cid:82) x − x + 34( x − x + 10)( x − dx ;c) Integrate trigonometric expressions: (cid:82) cos 9 x cos 2 xdx a) (cid:82) (cid:114) cos x sin x dx ;b) (cid:82) xdx
13 cos x + 4 sin x ;c) Integrate the functions with radicals: (cid:82) x √ − x dx ;a) (cid:82) ( √ x + 2 − dx ( x + 2)(1 + √ x + 2) .b) Solve the definite integrals: e − (cid:82) ln √ x + 1 dx ;a) π (cid:82) cos x sin xdx ;b) √ (cid:82) arctg x − x x dx ;c) (cid:82) dxx + √ x − . d) Find the area of the figure boundedby the curves: x + y = 6 , xy = 5 ;a) (cid:26) x = 3 cos t,y = 3 sin t ; x = 0 ( x ≥ ;b) ρ = 2 √ cos 2 φ. c) Find the arc-length of the curve: y = 4 − e x , ≤ x ≤ √ ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = 2 cos φ + 2 sin φ .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 3 x + 6 , − ≤ x ≤ , l = OX ;a) (cid:26) x = 2 + 4 cos t,y = 4 sin t ; ( x ≥ l = OY ;b) ρ = 3 e φ , ≤ φ ≤ π, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 4 x + 3 , y = − x + 7 , l = OX ;a) y = √ − x , y = x, x = 0 , l = OY ;b) ρ = 4(1 − cos φ ) , l = oρ .c) Solve the improper integrals: ∞ (cid:82) xe − x dx ;a) π/ (cid:82) sin 2 x √ − cos x dx .b)21 ersonal task 14 Integrate using the table andsubstitution under differential: (cid:82) √ x (1 + √ x ) dx ;a) (cid:82) sin x x dx ;b) (cid:82) x x +3 dx ;c) (cid:82) dx sh x + ch x ;d) Integrate the quadratic fractions: (cid:82) − x − x − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) − x − x + 4 x + 8 dx c) (cid:82) x dx √ x − x + 15 ;d) Integrate by parts or using thesuitable substitutions: (cid:82) x − x − dx ;a) (cid:82) dxx √ x − x − ;b) (cid:82) x (cos 2 x − sin 2 x ) dx ;c) (cid:82) ln( x − x − dx ;d) (cid:82) arcctg xdx ;e) (cid:82) e √ x − dx ;f) Integrate the polynomial fractions: (cid:82) x + 7 x − x − x + 2)( x + 2) dx ;a) (cid:82) x − x − x + 21 x − x + 3 x + 9 dx ;b) (cid:82) x + 64( x + 6 x + 10)( x − dx ;c) Integrate trigonometric expressions: (cid:82) cos 2 x sin 7 xdx a) (cid:82) cos x sin x dx ;b) (cid:82) dx x + 4 cos x ;c) Integrate the fractions with radicals: (cid:82) x √ − x dx ;a) (cid:82) √ x √ x − √ x dx .b) Solve the definite integrals: π/ (cid:82) e sin x sin 2 xdx ;a) π/ (cid:82) cos x sin xdx ;b) √ (cid:82) − x + 9 x + 1 dx ;c) ln 2 / (cid:82) e x √ e x − dx. d) Find the area of the figure boundedby the curves: y = ( x − , y = 3 x − ;a) (cid:26) x = 2 + 4 cos t,y = 2 sin t ; y = 1 ( y ≥ ;b) ρ = 3 sin 3 φ. c) Find the arc-length of the curve: y = 2 + ln(sin 2 x ) , π ≤ x ≤ π ;a) (cid:26) x = 2 cos t − cos 2 t,y = 2 sin t − sin 2 t ; 0 ≤ t ≤ π ;b) ρ = 4 φ, ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = ch x, ≤ x ≤ , l = OX ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π,l = OX ; b) ρ = 4 √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 4 x, y = − x, l = OX ;a) x = 6 − y , x = 2 , l = OY ;b) ρ = 2 cos φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) (4 x − dx x − x + 2 ;a) (cid:82) arcsin (cid:0) x (cid:1) − x √ − x dx .b)22 ersonal task 15 Integrate using the table andsubstitution under differential: (cid:82) (3 − x ) x dx ;a) (cid:82) (sin x − x ) dx ;b) (cid:82) x √ x + 9 dx ;c) (cid:82) x √ − x dx .d) Integrate the quadratic fractions: (cid:82) x + 7 x − dx ;a) (cid:82) dx √ x − x − ;b) (cid:82) ( − x + 11) dxx − x + 17 ;c) (cid:82) x + 3 √ − x − x dx .d) Integrate by parts or using thesuitable substitutions: (cid:82) dx √ e x − ;a) (cid:82) − dxx √ x + 4 x − ;b) (cid:82) x sin 3 xdx ;c) (cid:82) cos xe x dx ;d) (cid:82) ln(4 − x ) dx ;e) (cid:82) arctg √ xdx .f) Integrate the polynomial fractions: (cid:82) x − x + 28( x − x + 2)( x − dx ;a) (cid:82) x − x + 30 x − x − x + 3)( x − dx ;b) (cid:82) x + 8 x ( x + 4 x + 8)( x + 2) dx .c) Integrate trigonometric expressions: (cid:82) cos x sin x cos 4 xdx ;a) (cid:82) sin x cos xdx ;b) (cid:82)
12 cos xdx x − x + 9 .c) Integrate the fractions with radicals: (cid:82) √ − x x dx ;a) (cid:82) dx √ x + 3 + √ x + 3 .b) Solve the definite integrals: (cid:82) ( x − e − x dx ;a) π/ (cid:82) sin x cos x dx ;b) / (cid:82) ( − x + 5) dx √ x − x ;c) (cid:82) (cid:112) ( x − dx (cid:112) ( x − . d) Find the area of the figure boundedby the curves: x + 2 y = 7 , xy = 3 ;a) (cid:26) x = 2 cos t,y = 2 sin t ; y = 1 ( y ≥ ;b) ρ = 4(1 − sin φ ) .c) Find the arc-length of the curve: y = 4 − ch x, ≤ x ≤ ln 3 ;a) (cid:26) x = 2 cos t,y = 2 sin t ; 0 ≤ t ≤ π ;b) ρ = 5 √ e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = x , ≤ x ≤ , l = OY ;a) (cid:26) x = 2 + cos t,y = 3 + sin t ; l = OX ;b) ρ = 6 cos φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 5 , y = x + 5 , l = OX ;a) x = 4 − y , x = 8 − y , l = OY ;b) ρ = 4 sin 2 φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x + 1 x + x dx ;a) e (cid:82) dxx √ − ln x .b)23 ersonal task 16 Integrate using the table andsubstitution under differential: (cid:82) x e x + 7 e x e x dx ;a) (cid:82) sin 2 xdx (cos x + 3) − ;b) (cid:82) ln (3 − x )3 − x dx ;c) (cid:82) sin √ x √ x dx. d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) − dx √ x − x − ;b) (cid:82) (12 − x ) dxx − x + 18 ;c) (cid:82) − x √ − x − x dx .d) Integrate by parts or using thesuitable substitutions: (cid:82) √ e x + 4 dx ;a) (cid:82) dxx √ x − ;b) (cid:82) x (sin 2 x + cos 2 x ) dx ;c) (cid:82) (2 x − e x dx ;d) (cid:82) ( x + 4) ln xdx ;e) (cid:82) arcsin x √ x + 1 dx .f) Integrate the polynomial fractions: (cid:82) − x − x − x − x + 4)( x + 1) dx ;a) (cid:82) x + 10 x + 12 x − x + 4 x + 4 x dx ;b) (cid:82) x + 4 x + 36( x + 6 x + 18)( x − dx .c) Integrate trigonometric fractions: (cid:82) sin 5 x sin 3 xdx a) (cid:82) cos x sin xdx ;b) (cid:82) dx x + 7 cos x .c) Integrate the fractions with radicals: (cid:82) x dx √ − x ;a) (cid:82) (cid:114)(cid:16) x + 1 x − (cid:17) dx ( x + 1) .b) Solve the definite integrals: e +2 (cid:82) ln( x − dx ;a) π/ (cid:82) π/ x sin x dx ;b) √ (cid:82) − x √ x + 1 dx ;c) (cid:82) √ x −
45 + √ x − dx. d) Find the area of the figure boundedby the curves: y = 2 x − x + 16 , x + y = 6 ;a) (cid:26) x = 2 cos t,y = 2 sin t ; y = 0( y ≥ ;b) ρ = 2 √ cos 2 φ. c) Find the arc-length of the curve: y = ln( x + √ x − , ≤ x ≤ √ ;a) (cid:26) x = 4( t − sin t ) ,y = 4(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = 4 sin φ + 4 cos φ .c) Find the area of the surface formedby rotating the curves around the l -axis: x + y , l = OY ;a) (cid:26) x = e t cos t,y = e t sin t ; 0 ≤ t ≤ π, l = OX ;b) ρ = 6 sin φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 5 x + 1 , y = x + 1 , l = OX ;a) y = √ − x , y = x, x = 0 , l = OY ;b) ρ = 5(1 − cos φ ) , l = oρ .c) Solve the improper integrals: ∞ (cid:82) arctg (cid:0) x (cid:1) + 4 xx + 9 dx ;a) (cid:82) x + 1 √ x − x dx .b)24 ersonal task 17 Integrate using the table andsubstitution under differential: (cid:82) x ( √ x − dx ;a) (cid:82) dx sh x + 2 ch x ;b) (cid:82) x √ x − dx ;c) (cid:82) tg x cos x dx .d) Integrate the quadratic fractions: (cid:82) − x + 6 x − dx ;a) (cid:82) dx √ − x − x ;b) (cid:82) (4 x − dxx + 6 x + 10 ;c) (cid:82) (4 x − dx √ x + 2 x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) x + 7 √ x + 4 dx ;a) (cid:82) ln x − x (4 + ln x ) dx ;b) (cid:82) x sin 2 xdx ;c) (cid:82) x (2 x − x ) dx ;d) (cid:82) ln( x − √ x − dx ;e) (cid:82) x sin x cos x dx. f) Integrate the polynomial fractions: (cid:82) x + 19( x + x − x + 1) dx ;a) (cid:82) x + 3 x − x − x + x − x − dx ;b) (cid:82) x + 11 x + 11( x − x + 10)( x + 3) dx .c) Integrate trigonometric expressions: (cid:82) sin 2 x cos 5 xdx ;a) (cid:82) (cid:114) sin x cos x dx ;b) (cid:82) dx x −
12 cos x .c) Integrate the functions with radicals: (cid:82) √ x x dx ;a) (cid:82) dx √ x − √ x .b) Solve the definite integrals: ln 2 (cid:82) − (2 x + 2) e x dx ;a) π (cid:82) sin ( x ) dx ;b) / √ (cid:82) acrsin x − x √ − x dx ;c) π / (cid:82) cos √ xdx. d) Find the area of the figure boundedby the curves: y = ( x + 2) , y = 4 x + 8 ;a) (cid:26) x = 2 cos t,y = 5 sin t ; x = 1 ( x ≥ ;b) ρ = 4 sin 3 φ. c) Find the arc-length of the curve: y = ln(cos 4 x ) , − π ≤ x ≤ π ;a) (cid:26) x = 4 e − t cos t,y = 4 e − t sin t ; 0 ≤ t ≤ π ;b) ρ = 6 sin (cid:0) φ (cid:1) .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 1 + ch x, − ≤ x ≤ , l = OX ;a) (cid:26) x = 3( t − sin t ) ,y = 3(1 − cos t ); 0 ≤ t ≤ π,l = OX ;b) ρ = 2 √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 5 x + 1 , xy = 5 , l = OX ;a) y = 4 − x , y = 8 − x , l = OY ;b) ρ = 6 φ, ≤ φ ≤ π, l = oρ .c) Solve the improper integrals: ∞ (cid:82) dxx ( x + 1) ;a) (cid:82) (cid:16) (cid:17) x dxx .b)25 ersonal task 18 Integrate using the table andsubstitution under differential: (cid:82) x − x x x dx ;a) (cid:82) − x − cos 2 x dx ;b) (cid:82) ln (1 − x )1 − x dx ;c) (cid:82) ( cth x + th x ) dx .d) Integrate the quadratic fractions: (cid:82) − x − x − dx ;a) (cid:82) dx √ x − x − ;b) (cid:82) ( − x − dxx + 4 x + 13 ;c) (cid:82) ( − x + 10) dx √
15 + 2 x − x .d) Integrate by parts or using thesuitable substitutions: (cid:82) e x − e x √ − e x dx ;a) (cid:82) √ x + 4 x dx ;b) (cid:82) (2 x − x + 8) cos 2 xdx ;c) (cid:82) − x e x dx ;d) (cid:82) arccos √ x √ x dx e) (cid:82) x + 3) ln( x + 9) dx .f) Integrate the polynomial fractions: (cid:82) x − x + 20( x − x + 6)( x + 1) dx ;a) (cid:82) x + x + 5 x + 2 x − x − x + 1 dx ;b) (cid:82) x − x + 13 x − x + 13 x dx .c) Integrate trigonometric expressions: (cid:82) sin 5 x sin 3 xdx ;a) (cid:82) cos x sin x dx ;b) (cid:82) dx x .c) Integrate the fractions with radicals: (cid:82) x + 2) − dx √ x + x ;a) (cid:82) ( √ x + 1) dx √ x + 4 √ x .b) Solve the definite integrals: e − (cid:82) ln( x + 1)( x + 1) dx ;a) π (cid:82) cos x sin xdx ;b) (cid:82) x − √ x + 2 dx ;c) π/ (cid:82) ( e − tg x + sin x ) dx cos x . d) Find the area of the figure boundedby the curves: y = ( x + 1) , y = 5( x + 1) ;a) (cid:26) x = 3 cos t,y = 4 sin t ; y = 2 ( y ≥ ;b) ρ = 4 √ sin 2 φ. c) Find the arc-length of the curve: y = 2 − ch x, ≤ x ≤ ln 3 ;a) (cid:26) x = 3( t − sin t ) ,y = 3(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = √ e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = x , − ≤ x ≤ , l = OX ;a) (cid:26) x = 3 + cos t,y = 2 + sin t ; l = OY ;b) ρ = 6(1 + cos φ ) , l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 6 x, y = − x + 12 x, l = OX ;a) y = x , y = 3 x − , l = OY ;b) ρ = 8 cos φ, l = oρ .c) Solve the improper integrals: (cid:82) −∞ x e − x dx ;a) (cid:82) x − √ x − x dx .b)26 ersonal task 19 Integrate using the table andsubstitution under differential: (cid:82) (1 − √ x ) dx √ x ;a) (cid:82) ctg x − cos x dx ;b) (cid:82) x − x √ − x dx ;c) (cid:82) (cos x + 3cos x ) dx. d) Integrate the quadratic fractions: (cid:82) − x + 6 x − dx ;a) (cid:82) dx √ − x − x ;b) (cid:82) (4 − x ) dxx + 4 x + 20 ;c) (cid:82) (6 x − dx √ x − x + 5 .d) Integrate by parts or using thesuitable substitutions: (cid:82) − x ( x + 2) dx ;a) (cid:82) (2 + 4 ln x ) dxx (cid:112) ln x + 4 ;b) (cid:82) x sin 4 xdx ;c) (cid:82) x − xe x dx ;d) (cid:82) ln( √ x + 1) dx ;e) (cid:82) x + 4sin x dx .f) Integrate the polynomial fractions: (cid:82) x + 24 x − x + 2 x − x dx ;a) (cid:82) x + 17 x + 27 x + 16( x + 3 x + 2)( x + 1) dx ;b) (cid:82) x + 14 x − x − x + 5)( x + 2) dx .c) Integrate trigonometric expressions: (cid:82) cos x sin 9 xdx ;a) (cid:82) cos x sin x dx ;b) (cid:82) dx x + 5 cos x .c) Integrate the fractions with radicals: (cid:82) dxx √ − x ;a) (cid:82) (cid:114) x + 2 x − dx ( x + 2) .b) Solve the definite integrals: π / (cid:82) π / (1 − sin √ x ) dx √ x ;a) π/ (cid:82) x cos xdx ;b) √ (cid:82) x + 4 x + 1 dx ;c) (cid:82) x (cid:112) (9 − x ) dx. d) Find the area of the figure boundedby the curves: y = − x + 4 x + 1 , xy = 4 ;a) (cid:26) x = 2 cos t,y = 2 + 4 sin t ; x = 1 ( x ≥ ;b) ρ = 3 cos 3 φ. c) Find the arc-length of the curve: y = 2 − e x , ln 2 ≤ x ≤ ln 6 ;a) (cid:26) x = 3(2 cos t − cos 2 t ) ,y = 3(2 sin t − sin 2 t ); 0 ≤ t ≤ π ;b) ρ = 2 cos φ + 2 sin φ .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 3 x + 9 , − ≤ x ≤ , l = OX ;a) (cid:26) x = 4 cos t,y = 4 sin t ; l = OY ;b) ρ = (cid:113) sin 2 (cid:0) φ − π (cid:1) , l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = ( x − , y = x + 4 , l = OX ;a) y = x − , y = 2 x, x = 0 ( x ≥ ,l = OY ;b) ρ = φ, ≤ φ ≤ π, l = oρ .c) Solve the improper integrals: ∞ (cid:82) − / arctg (cid:0) x (cid:1) x dx ;a) (cid:82) (2 x + 1) dx √ x − x − .b)27 ersonal task 20 Integrate using the table andsubstitution under differential: (cid:82) √ x (2 − x √ x ) dx ;a) (cid:82) x − cos 2 x dx ;b) (cid:82) − x √ − x dx ;c) (cid:82) (1 − x dx .d) Integrate the quadratic fractions: (cid:82) − xx − dx ;a) (cid:82) dx √
12 + 4 x − x ;b) (cid:82) − xdxx − x + 5 ;c) (cid:82) (14 − x ) dx √ x − x + 25 .d) Integrate by parts or using thesuitable substitutions: (cid:82) (1 − e x ) e x dx ;a) (cid:82) dx ( x − √ x − x ;b) (cid:82) (4 − x ) cos 4 xdx ;c) (cid:82) ln ( x + 2) dx ;d) (cid:82) √ x +2 dx ;e) (cid:82) arccos( x − dx .f) Integrate the polynomial fractions: (cid:82) x − x − x + x − x − dx ;a) (cid:82) x − x + x + 22 x − x − x − dx ;b) (cid:82) x + 10 x + 26( x + 6 x + 13)( x − dx .c) Integrate trigonometric expressions: (cid:82) cos 2 x cos 5 xdx ;a) (cid:82) √ cos x sin xdx ;b) (cid:82) dx x − x cos x − x .c) Integrate the fractions with radicals: (cid:82) x dx (cid:112) (4 − x ) ;a) (cid:82) xdx ( x + 5) √ x .b) Solve the definite integrals: ln 2+1 (cid:82) ( x − e x − dx ;a) π (cid:82) sin x cos xdx ;b) / (cid:82) (2 x + 7) dx √ x − x ;c) √ (cid:82) − √ x + 1 − √ x + 1 + 2 dx. d) Find the area of the figure boundedby the curves: x + y = 7 , xy = 3 ;a) (cid:26) x = 4 cos t,y = 2 + 4 sin t ; x = 2 ( x ≥ ;b) ρ = 4(1 − sin φ ) .c) Find the arc-length of the curve: y = 4 ln(16 − x ) , − ≤ x ≤ ;a) (cid:26) x = 3 cos t,y = 3 sin t ; 0 ≤ t ≤ π ;b) ρ = φ, ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 4 ch (cid:0) x (cid:1) , ≤ x ≤ , l = OX ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π,l = OX ;b) ρ = 8 cos φ, l = oρ. c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x, y = 2 x − x, l = OX ;a) y = 9 − x , x = 8 , l = OY ;b) ρ = 4 sin 2 φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) − x + 5 x + 1 dx ;a) π/ (cid:82) x √ − sin x dx .b)28 ersonal task 21 Integrate using the table andsubstitution under differential: (cid:82) (1 − x ) x dx ;a) (cid:82) − sin x sin x dx ;b) (cid:82) x e x +5 dx ;c) (cid:82) (sin 2 x − x ) dx cos x + 9 .d) Integrate the quadratic fractions: (cid:82) − xx − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) (6 x + 7) dxx + 4 x + 5 c) (cid:82) (2 x − dx √ x − x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) √ e x + 16 dx ;a) (cid:82) dxx √ − x ;b) (cid:82) ( x −
6) sin xdx ;c) (cid:82) x ln( x + 1) dx ;d) (cid:82) arcsin √ xdx ;e) (cid:82) cos( ln x ) dx .f) Integrate the polynomial fractions: (cid:82) x + 5 x − x − x − x − dx ;a) (cid:82) − x − x + x + 20 x + 3 x − dx ;b) (cid:82) − x − x + 44( x − x + 20)( x + 2) .c) Integrate trigonometric expressions: (cid:82) sin 5 x cos 2 xdx ;a) (cid:82) cos x sin xdx ;b) (cid:82) − xdx − sin x + 3 cos x .c) Integrate the fractions with radicals: (cid:82) √ x + 9 x dx ;a) (cid:82) (4 − √ x ) dx (2 + √ x ) √ x .b) Solve the definite integrals: ln 2+1 (cid:82) ( x − e x − dx ;a) π/ (cid:82) sin x cos x dx ;b) √ (cid:82) x √ x − dx ;c) (cid:82) √ x + 1 dx (cid:112) ( x + 1) . d) Find the area of the figure boundedby the curves: y = 2 x − x + 10 , y = − x + 3 x + 1 ;a) (cid:26) x = 2 cos t,y = 4 sin t ; x = 1 ( x ≥ ;b) ρ = 2 cos 4 φ. c) Find the arc-length of the curve: y = 2 + ln x, ≤ x ≤ √ ;a) (cid:26) x = 3( t − sin t ) ,y = 3(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = √ e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = √ x x − , ≤ x ≤ , l = OX ;a) (cid:26) x = 1 + 2 cos t,y = 2 sin t ; ( x ≥ l = OY ;b) ρ = √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 10 , y = x + 10 , l = OX ;a) x = − y + 6 y − , x = 0 , l = OY ;b) ρ = 4(1 + cos φ ) , l = oρ .c) Solve the improper integrals: ∞ (cid:82) x + arcctg xx + 1 dx ;a) / (cid:82) − x + 3 √ − x dx .b)29 ersonal task 22 Integrate using the table andsubstitution under differential: (cid:82) (2 + √ x ) dxx ;a) (cid:82) dx sh x − ch x ;b) (cid:82) x x − dx ;c) (cid:82) sin 2 x √ − cos x dx .d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) (4 x − dxx − x + 13 ;c) (cid:82) ( − x − dx √ x + 6 x + 5 .d) Integrate by parts or using thesuitable substitutions: (cid:82) (3 x − dx ( x + 1) ;a) (cid:82) dx ( x − √ x − x ;b) (cid:82) x sin 2 xdx ;c) (cid:82) e x (cos 4 x + 2) dx d) (cid:82) x ln xdx ;e) (cid:82) arctg √ x √ x dx .f) Integrate the polynomial fractions: (cid:82) x + 20 x + 24( x − x − x + 2) dx ;a) (cid:82) − x + 3 x + 4 x − x − x + x dx ;b) (cid:82) x − x − x + 2 x + 2)( x − dx .c) Integrate trigonometric expressions: (cid:82) sin x sin 7 xdx ;a) (cid:82) cos x sin x dx ;b) (cid:82) dx − x + 4 cos x .c) Integrate the fractions with radicals: (cid:82) x dx (cid:112) (9 − x ) ;a) (cid:82) (cid:114) x + 1 x − dx ( x + 1) .b) Solve the definite integrals: log (cid:82) x x dx ;a) π (cid:82) cos ( x ) sin ( x ) dx ;b) √ e − (cid:82) − x x + 1 dx ;c) e (cid:82) ln xdxx √ x . d) Find the area of the figure boundedby the curves: y = ( x + 2) , y = 4 x + 8 ;a) (cid:26) x = 4 cos t,y = 3 sin t ; x = 2 ( x ≥ ;b) ρ = 3 √ cos 2 φ. c) Find the arc-length of the curve: y = √ − x , −√ ≤ x ≤ ;a) (cid:26) x = 3 e − t cos t,y = 3 e − t sin t ; 0 ≤ t ≤ π ;b) ρ = 6(1 + cos φ ) .c) Find the area of the surface formedby rotating the curves around the l -axis: y = √ x + 12 , − ≤ x ≤ , l = OX ;a) (cid:26) x = cos t,y = sin t ; l = OY ;b) ρ = 2 sin φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x + 4 x + 4 , y = 4 − x, l = OX ;a) x = − y + 3 y, x = − y + 6 y, l = OY ;b) ρ = sin 2 φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) − x x − dx ;a) (cid:82) − ln 2 e x √ − e x dx .b)30 ersonal task 23 Integrate using the table andsubstitution under differential: (cid:82) (3 − x ) x dx ;a) (cid:82) (3 + ctg x ) dx ;b) (cid:82) ln (5 − x )5 − x dx ;c) (cid:82) cos (2 x + 5) dx .d) Integrate the quadratic fractions: (cid:82) x + 5 x − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) xdxx + 2 x + 10 ;c) (cid:82) (4 x + 4) dx √ − x − x .d) Integrate by parts or using thesuitable substitutions: (cid:82) √ x + 3 dx √ x + 3 ;a) (cid:82) xdxx √ x ;b) (cid:82) x − xe x dx ;c) (cid:82) (8 − x ) sin 4 xdx ;d) (cid:82) x cos xdx sin x .e) (cid:82) arccos(1 − x ) dx ;f) Integrate the polynomial fractions: (cid:82) x + 10 x − x − x + 10)( x + 1) dx ;a) (cid:82) − x + 2 x + 15 x − x − x + 4 x dx ;b) (cid:82) x + 19 x − x − x + 10)( x + 3) dx .c) Integrate trigonometric expressions: (cid:82) cos 2 x cos 6 xdx ;a) (cid:82) cos x sin xdx ;b) (cid:82) − dx x − x cos x − x .c) Integrate the fractions with radicals: (cid:82) dxx √ x − ;a) (cid:82) dx √ x + 4 √ x .b) Solve the definite integrals: e +1 (cid:82) ln ( x − dx ;a) π/ (cid:82) π/ cos x sin x dx ;b) √ (cid:82) x dx x ;c) (cid:82) dxx + √ x + 4 . d) Find the area of the figure boundedby the curves: x + y = 9 , xy = 8 ;a) (cid:26) x = 2 cos t,y = 2 sin t ; x = 1 ( x ≥ ;b) ρ = 4 sin 2 φ. c) Find the arc-length of the curve: y = ln( x + √ x − , ≤ x ≤ √ ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = 3 √ e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 2 ch (cid:0) x (cid:1) , − ≤ x ≤ , l = OX ;a) (cid:26) x = 3 + cos t,y = 1 + sin t ; l = OY ;b) ρ = 4 √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 10 , y = x + 10 , l = OX ;a) y = ln x, x = 0 , y = 0 , y = ln 5 , l = OY ;b) ρ = 2(1 − cos φ ) , l = oρ .c) Solve the improper integrals: ∞ (cid:82) xdx (cid:112) ( x + 1) ;a) π / (cid:82) sin √ x − √ x dx .b)31 ersonal task 24 Integrate using the table andsubstitution under differential: (cid:82) x (4 − x ) dx ;a) (cid:82) sin 2 xdx − (sin x − ;b) (cid:82) x x − dx ;c) (cid:82) th xdx .d) Integrate the quadratic fractions: (cid:82) − x − x − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) (2 x + 6) dxx − x + 5 ;c) (cid:82) (3 x − dx √ x + 4 x − .d) Integrate by parts of using thesuitable substitutions: (cid:82) ( e x + 1) √ e x + 1 dx ;a) (cid:82) dx ( x − √ x − x ;b) (cid:82) (4 − x ) cos 2 x sin 2 xdx ;c) (cid:82) (2 − x )3 x dx ;d) (cid:82) x √ x dx ;e) (cid:82) (5 x − arctg x dx .f) Integrate the polynomial fractions: (cid:82) − x + 3 x − x − x + 2)( x − dx ;a) (cid:82) x − x − x − x − x − dx ;b) (cid:82) x − x − x + 4 x + 20 x dx .c) Integrate trigonometric expressions: (cid:82) cos 3 x sin 9 xdx ;a) (cid:82) cos x sin x dx ;b) (cid:82) dx − x + 5 cos x .c) Integrate the fractions with radicals: (cid:82) √ − x x dx ;a) (cid:82) √ x + √ x √ x dx .b) Solve the definite integrals: (cid:82) (4 x − e − x dx ;a) π/ (cid:82) cos x sin xdx ;b) (cid:82) x − x − dx ;c) (cid:82) − x √ − x dx. d) Find the area of the figure boundedby the curves: y = ( x − , y = 3 x − ;a) (cid:26) x = t − sin t,y = 1 − cos t ; y = 0 , ≤ t ≤ π ;b) ρ = 3 cos 3 φ. c) Find the arc-length of the curve: y = + ln(sin 3 x ) , π ≤ x ≤ π ;a) (cid:26) x = 2(2 cos t − cos 2 t ) ,y = 2(2 sin t − sin 2 t ); 0 ≤ t ≤ π ;b) ρ = 4 φ, ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: x + y
16 = 1 , l = OY ;a) (cid:26) x = e t cos t,y = e t sin t ; 0 ≤ t ≤ π, l = OX ;b) ρ = 4 (cid:113) sin 2 (cid:0) φ − π (cid:1) , l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 6 x + 1 , xy = 6 , l = OX ;a) x = 4 y − y , x = 8 y − y , l = OY ;b) ρ = 4 sin φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) − − x − x − x + 5 dx ;a) (cid:82) / √ dxx √ − x .b)32 ersonal task 25 Integrate using the table andsubstitution under differential: (cid:82) (3 x e x − e x ) dxe x ;a) (cid:82) − tg x tg x dx ;b) (cid:82) x dx √ x + 9 ;c) (cid:82) sin 2 x − x cos x − dx .d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) dx √ − x − x ;b) (cid:82) ( − x + 4) dxx − x + 12 ;c) (cid:82) (4 x − dx √ x − x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) √ x + 2 x − dx ;a) (cid:82) dxx √ x + 1 ;b) (cid:82) (2 − x − x ) cos 2 xdx ;c) (cid:82) sin 3 xe x dx ;d) (cid:82) ln( x − x − dx ;e) (cid:82) arcsin x √ x + 4 dx .f) Integrate the polynomial fractions: (cid:82) x + 14( x + 3 x + 2)( x − dx ;a) (cid:82) x − x − x + 4 x + 2 x − x − dx ;b) (cid:82) x − x − x + 6 x + 18)( x − dx .c) Integrate trigonometric expressions: (cid:82) sin 6 x sin 4 xdx ;a) (cid:82) cos x sin xdx ;b) (cid:82) xdx x + 6 cos x .c) Integrate the functions with radicals: (cid:82) x √ − x dx ;a) (cid:82) (cid:114) x + 4 x − dx ( x + 4) .b) Solve the definite integrals: (cid:82) ( x − e x dx ;a) π/ (cid:82) sin x cos x dx ;b) (cid:82) − xdx √ − x − x ;c) (cid:82) x + 3 √ x − dx. d) Find the area of the figure boundedby the curves: y = 2 x − x + 10 , y = − x + 3 x + 1 ;a) (cid:26) x = 6 cos t,y = 2 sin t ; x = 0 ( x ≥ ;b) ρ = 2(1 + sin φ ) .c) Find the arc-length of the curve: y = 2 − ch x, − ln 2 ≤ x ≤ ln 2 ;a) (cid:26) x = cos t,y = sin t ; 0 ≤ t ≤ π ;b) ρ = √ e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = x , ≤ x ≤ √ , l = OY ;a) (cid:26) x = 2( t − sin t ) ,y = 2(1 − cos t ); 0 ≤ t ≤ π,l = OX ; b) ρ = 8 cos φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = ( x − , y = 16 x − , ( x ≥ l = OX ;a) x = 8 + y , x = 16 − y , l = OY ;b) ρ = sin 2 φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) −√ x − xx + 4 dx ;a) e π (cid:82) sin(ln x ) dxx .b)33 ersonal task 26 Integrate using the table andsubstitution under differential: (cid:82) ( x − x √ x dx a) (cid:82) (4 − ctg x ) dx ;b) (cid:82) x e x − dx ;c) (cid:82) √ x · dx √ x .d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) dx √
15 + 4 x − x ;b) (cid:82) ( − x + 14) dxx − x + 13 ;c) (cid:82) x + 6 √ x + 6 x dx .d) Integrate by parts or using thesuitable substitutions: (cid:82) (2 ln x + 8) dxx (cid:112) − ln x ;a) (cid:82) dx ( x − √ x − x ;b) (cid:82) x sin 4 xdx ;c) (cid:82) x − x + 12 e x dx ;d) (cid:82) ( x + 3) ln xdx ;e) (cid:82) (arccos 2 x ) dx .f) Integrate the polynomial fractions: (cid:82) x − x + 12 x − x − x + 4 dx ;a) (cid:82) x − x − x + 6 x − x − x − dx ;b) (cid:82) x − x ( x − x + 8)( x − dx .c) Integrate trigonometric fractions: (cid:82) sin 7 x cos 4 xdx a) (cid:82) cos x sin xdx ;b) (cid:82) − dx x + 6 sin x cos x − x .c) Integrate the fractions with radicals: (cid:82) √ x − x dx ;a) (cid:82) √ x + 2 √ x + √ x dx .b) Solve the definite integrals: e − (cid:82) − ln ( x + 2) √ x + 2 dx ;a) π/ (cid:82) π/ cos x sin x dx ;b) (cid:82) x + 2 xx + 4 dx ;c) (cid:82) √ x − dx. d) Find the area of the figure boundedby the curves: x + y = 7 , xy = 3 ;a) (cid:26) x = cos t,y = 2 sin t ; x = 0 ( x ≥ ;b) ρ = 3 sin 4 φ. c) Find the arc-length of the curve: y = ln( x + √ x − , ≤ x ≤ ;a) (cid:26) x = 4( t − sin t ) ,y = 4(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = 2 cos φ + 2 sin φ .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 2 x + 6 , ≤ x ≤ , l = OX ;a) (cid:26) x = 3 + 2 cos t,y = 1 + 2 sin t ; l = OY ;b) ρ = √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x + 4 x + 6 , y = x + 6 , l = OX ;a) y = √ − x , y = x, y = 0 , l = OY ;b) ρ = 8 sin (cid:16) φ (cid:17) , l = oρ .c) Solve the improper integrals: (cid:82) −∞ e x cos xdx ;a) (cid:82) dxx √ x − .b)34 ersonal task 27 Integrate using the table andsubstitution under differential: (cid:82) (2 x − √ x dx ;a) (cid:82) cos 2 x − x + 1 dx ;b) (cid:82) x √ x − dx ;c) (cid:82) (cid:16) − tg x (cid:17) dx ;d) Integrate the quadratic fractions: (cid:82) x + 12 x − dx ;a) (cid:82) dx √ − x − x ;b) (cid:82) ( − x + 16) dxx − x + 20 ;c) (cid:82) xdx √ x + 4 x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) dx √ e x − ;a) (cid:82) √ x x dx ;b) (cid:82) x (sin x + cos x ) dx ;c) (cid:82) x x dx ;d) (cid:82) arctg √ x · dx √ x ;e) (cid:82) ln( x − dx .f) Integrate the polynomial fractions: (cid:82) x − x + 6( x − x + 3)( x + 2) dx ;a) (cid:82) − x − x − x − x + 4 x + 4 x dx ;b) (cid:82) x + 26 x − x − x + 5)( x + 5) dx .c) Integrate trigonometric expressions: (cid:82) sin x cos 5 xdx ;a) (cid:82) cos x √ sin xdx ;b) (cid:82) dx
13 + 12 sin x .c) Integrate the fractions with radicals: (cid:82) √ x − x dx ;a) (cid:82) (cid:114) x − x + 1 dx ( x − .b) Solve the definite integrals: (cid:82) ( x − e x dx ;a) π (cid:82) cos ( x ) sin ( x ) dx ;b) √ (cid:82) x + 8 x √ − x dx ;c) e − (cid:82) ln x √ x + 1 dx. d) Find the area of the figure boundedby the curves: y = ( x − , y = 6 x − ;a) (cid:26) x = 6 cos t,y = 5 sin t ; x = 3 ( x ≥ ;b) ρ = 4 √ cos 2 φ. c) Find the arc-length of the curve: y = ln(cos 3 x ) , − π ≤ x ≤ π ;a) (cid:26) x = 3 cos t,y = 3 sin t ; x = 0 ( x ≥ ;b) ρ = 8 cos (cid:16) φ (cid:17) .c) Find the area of the surface formedby rotating the curves around the l -axis: x + y = 4 , l = OY ;a) (cid:26) x = t − sin t,y = 1 − cos t ; 0 ≤ t ≤ π, l = OX ;b) ρ = 6 sin φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 4 x + 1 , xy = 4 , l = OX ;a) y = 4 − x , y = 12 − x , l = OY ;b) ρ = 3 e − φ , ≤ φ ≤ π, l = oρ .c) Solve the improper integrals: ∞ (cid:82) dxx + √ x ;a) (cid:82) x √ − x dx .b)35 ersonal task 28 Integrate using the table andsubstitution under differential: (cid:82) x − x x x dx ;a) (cid:82) ( tg x − ctg x ) dx ;b) (cid:82) arcsin x − x √ − x dx ;c) (cid:82) cos x + sin x sin x − cos x dx .d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) dx √ − x − x ;b) (cid:82) ( − x + 5) dxx − x + 2 ;c) (cid:82) ( − x − dx √ x − x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) ln xdxx √ x ;a) (cid:82) √ x − dx √ x − ;b) (cid:82) x cos xdx ;c) (cid:82) e x sin 3 xdx ;d) (cid:82) x arccos x √ − x dx ;e) (cid:82) (2 x + 1) sin xdx cos x .f) Integrate the polynomial fractions: (cid:82) − x + 10 x − x − x − x + 4 dx ;a) (cid:82) x − x − x + 38( x + 4 x − x − dx ;b) (cid:82) x − x + 4 x + 13)( x + 1) dx .c) Integrate trigonometric expressions: (cid:82) sin 8 x sin 4 xdx ;a) (cid:82) (cid:114) sin x cos x dx ;b) (cid:82) xdx x + 13 cos x .c) Integrate the fractions with radicals: (cid:82) x dx (cid:112) (9 − x ) ;a) (cid:82) dx √ x + 1 + 4 √ x + 1 .b) Solve the definite integrals: e (cid:82) ln xdx ;a) π (cid:82) cos x sin xdx ;b) (cid:82) √ xdxx + 4 ;c) (cid:82) √ x √ x − dx. d) Find the area of the figure boundedby the curves: y = ( x + 2) , y = 9( x + 2) ;a) (cid:26) x = 4 cos t,y = 2 sin t ; y = 1 ( y ≥ ;b) ρ = 6(1 − sin φ ) . c) Find the arc-length of the curve: y = 2 − e x , ≤ x ≤ ln 15 ;a) (cid:26) x = 3( t − sin t ) ,y = 3(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = √ e − φ , ≤ φ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 4 x , ≤ x ≤ , l = OY ;a) (cid:26) x = 2 + cos t,y = 1 + sin t ; l = OX ;b) ρ = 3 √ cos 2 φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = − x + 5 x + 1 , y = 6 − x, l = OX ;a) y = 1 − x , y = 3 − x , l = OY ;b) ρ = 2 sin 2 φ, l = oρ .c) Solve the improper integrals: ∞ (cid:82) arctg x x dx ;a) π / (cid:82) cos √ x √ x dx .b)36 ersonal task 29 Integrate using the table andsubstitution under differential: (cid:82) (2 + 3 √ x ) x dx ;a) (cid:82) tg xdx ;b) (cid:82) x (cid:112) (1 − x ) dx ;c) (cid:82) ( th x + cth x ) dx. d) Integrate the quadratic fractions: (cid:82) x − x − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) (4 x − dxx − x + 18 ;c) (cid:82) (6 x − dx √ − x + x .d) Integrate by parts or using thesuitable substitutions: (cid:82) x + 6( x + 1) dx ;a) (cid:82) dx ( x − √ x − x ;b) (cid:82) (4 x − x ) sin x cos xdx ;c) (cid:82) e x sin 3 xdx ;d) (cid:82) ln xx dx ;e) (cid:82) (3 x + 1) arcctg xdx .f) Integrate the polynomial fractions: (cid:82) − x + 15 x − x − x + 6)( x + 1) dx ;a) (cid:82) x − x − x − x + x − x − dx ;b) (cid:82) − x + 80( x − x + 10)( x + 4) dx .c) Integrate trigonometric expressions: (cid:82) cos 2 x sin 8 xdx ;a) (cid:82) cos x sin x dx ;b) (cid:82) dx − x + 6 cos x .c) Integrate the fractions with radicals: (cid:82) √ x x dx ;a) (cid:82) (cid:114) x + 3 x − dx ( x + 3) .b) Solve the definite integrals: / ln 2 (cid:82) / ln 4 x √ e · dxx ;a) π (cid:82) cos x sin xdx ;b) √ (cid:82) x + 4 x + 1 dx ;c) (cid:82) x √ − x dx. d) Find the area of the figurebounded by the curves: y = 2 x − x + 13 , y = − x + 4 x + 1 ;a) (cid:26) x = 4 cos t,y = 2 + 4 sin t ; x = 2 ( x ≥ ;b) ρ = 3 √ sin 2 φ. c) Find the arc-length of the curve: y = 4 ln(16 − x ) , − ≤ x ≤ ;a) (cid:26) x = 2 e − t cos t,y = 2 e − t sin t ; 0 ≤ t ≤ π ;b) ρ = 4 cos (cid:16) φ (cid:17) .c) Find the area of the surface formedby rotating the curves around the l -axis: y = ch x, ≤ x ≤ , l = OX ;a) (cid:26) x = 3 cos t,y = 3 sin t ; l = OY ;b) ρ = 2 cos φ, l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x + 4 , y = 6 x − , l = OX ;a) y = x − , y = 5 x, x = 0 ( x ≥ , l = OY ;b) ρ = 6 φ, ≤ φ ≤ π, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x e − x dx ;a) (cid:82) x − x √ − x dx .b)37 ersonal task 30 Integrate using the table andsubstitution under differential: (cid:82) √ x (1 + 2 x √ x ) dx ;a) (cid:82) cos 2 x + 1sin x dx ;b) (cid:82) arccos x − x √ − x dx ;c) (cid:82) (cid:16) sh ( x + 1) − (cid:17) dx .d) Integrate the quadratic fractions: (cid:82) − x + 6 x − dx ;a) (cid:82) dx √ x − x − ;b) (cid:82) ( − x + 4) dxx + 2 x + 17 ;c) (cid:82) − x + 6 √ x − x dx .d) Integrate by parts or using thesuitable substitutions: (cid:82) √ x − dxx + 4 √ x − ;a) (cid:82) dxx √ − x ;b) (cid:82) x cos 3 xdx ;c) (cid:82) x ln( x − dx ;d) (cid:82) e √ x +4 dx ;e) (cid:82) x · arcsin ( x ) dx .f) Integrate the polynomial fractions: (cid:82) x + 5 x − x + 3 x + 2)( x + 3) dx ;a) (cid:82) − x + 4 x + 6 x − x − x + 8)( x − dx ;b) (cid:82) x − x − x + 2 x + 5)( x − dx .c) Integrate trigonometric expressions: (cid:82) sin 5 x cos 3 xdx ;a) (cid:82) sin x cos xdx ;b) (cid:82) dx x −
12 sin x cos x + 5 cos x .c) Integrate the fractions with radicals: (cid:82) x √ x + 4 dx ;a) (cid:82) ( √ x + 1) dx √ x + 2 √ x − .b) Solve the definite integrals: (cid:82) arctg xdx ;a) π/ (cid:82) sin x cos x dx ;b) (cid:82) x + 2 √ x − x dx ;c) (cid:82) − x √ − x dx. d) Find the area of the figure boundedby the curves: y = ( x + 2) , y = 4 x + 8 ;a) (cid:26) x = 4 cos t,y = 6 sin t ; x = 2 ( x ≥ ;b) ρ = 3 cos 2 φ .c) Find the arc-length of the curve: y = 1 − ch x, ≤ x ≤ ln 2 ;a) (cid:26) x = 4 cos t,y = 4 sin t ; 0 ≤ t ≤ π ;b) ρ = 6 cos φ + 6 sin φ .c) Find the area of the surface formedby rotating the curves around the l -axis: y = √ x + 8 , − ≤ x ≤ , l = OX ;a) (cid:26) x = 3( t − sin t ) ,y = 3(1 − cos t ); 0 ≤ t ≤ π,l = OX ; b) ρ = 2(1 + cos φ ) , l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x − x, y = 2 x − x, l = OX ;a) x = 9 − y , x + 3 y = 9 , l = OY ;b) ρ = e − φ , ≤ φ ≤ π, l = oρ .c) Solve the improper integrals: ∞ (cid:82) x + 5 x + 1 dx ;a) e (cid:82) dxx √ − ln x .b)38 ample task Integrate using the table andsubstitution under differential: (cid:82) x (1 − x ) dx ;a) (cid:82) (3 + tg x ) dx ;b) (cid:82) x e x − dx ;c) (cid:82) sin (1 + 3 x ) dx .d) Integrate the quadratic fractions: (cid:82) − x + 5 x − dx ;a) (cid:82) dx √ x − x ;b) (cid:82) ( − x − dxx + 6 x + 10 ;c) (cid:82) (8 x − dx √ x + 4 x − .d) Integrate by parts or using thesuitable substitutions: (cid:82) x + 2 √ x + 4 dx ;a) (cid:82) dxx √ x − x + 1 ;b) (cid:82) (4 x − e x dx ;c) (cid:82) x ln xdx ;d) (cid:82) sin √ x + 1 dx ;e) (cid:82) x sin x cos x dx .f) Integrate the polynomial fractions: (cid:82) − x + 2 x + 13 x + 2 x − x − dx ;a) (cid:82) x − x + 56 x − x − x + 8 dx ;b) (cid:82) x − x − x + 4 x + 5)( x − dx .c) Integrate trigonometric functions: (cid:82) sin 10 x sin 3 xdx ;a) (cid:82) (cid:114) sin x cos x dx ;b) (cid:82) dx x + 3 sin x + 6 .c) Integrate the fractions with radicals: (cid:82) dxx √ − x ;a) (cid:82) (cid:114) x + 1 x − dx ( x − .b) Solve the definite integrals: (cid:82) ln(1 + x ) dx ;a) π/ (cid:82) sin x √ cos xdx ;b) √ (cid:82) x dx √ − x ;c) (cid:82) √ x √ x + √ x dx. d) Find the area of the figurebounded by the curves: y = 2 x − x + 6 , y = x − x ;a) (cid:26) x = 4 cos t,y = 4 sin t ; y = ( y ≥ ) ;b) ρ = 4 sin 3 ϕ. c) Find the arc length of the curves: y = 13 ln(cos 3 x ) , ≤ x ≤ π ;a) (cid:26) x = 4( t − sin t ) ,y = 4(1 − cos t ); 0 ≤ t ≤ π ;b) ρ = √ e ϕ , ≤ ϕ ≤ π .c) Find the area of the surface formedby rotating the curves around the l -axis: y = 12 ch 2 x, − ≤ x ≤ , l = OX ;a) (cid:26) x = 3 + cos t,y = 2 + sin t ; l = OY ;b) ρ = √ cos 2 φ, ≤ φ ≤ π l = oρ .c) Find the volume of the body formedby rotating the curves around the l -axis: y = x + 2 x + 5 , y = 5 − x, l = OX ;a) x = 5 + 4 y − y , x = 5 , l = OY ;b) ρ = 6(1 + cos ϕ ) , l = oρ .c) Solve the improper integrals: ∞ (cid:82) − x − x + 4 dx ;a) π/ (cid:82) e − tg x dx cos x .b)39 .(a) ˙ I = (cid:90) x (1 − x ) dx = (cid:12)(cid:12)(cid:12)(cid:12) Applying the Newton binomto the function under the integral (cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) x (1 − x + x ) dx = (cid:90) ( x − x + x ) dx = 14 x − x + 18 x + C == 14 x − x + 18 x + C. Result: ˙ I = 14 x − x + 18 x + C, C ∈ R . ∗ Solution of Problem 1(a)guided by Ricard Riba isavailable on-line: https://youtu.be/x48CikKlF9c ˙ I = (cid:90) (3 + tg x ) dx = (cid:90) (cid:0) x ) (cid:1) dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Applying thetrigonometric identity x = 1cos x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) (cid:18) x (cid:19) dx = 2 x + (cid:90) dx cos x = 2 x + tg x + C. Result: ˙ I = 2 x + tg x + C, C ∈ R . ∗ Solution of Problem 1(b)guided by Irina Blazhievskais available on-line: https://youtu.be/1MqmZbQ-3qM ˙ I = (cid:90) x e x − dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Checking the exact differential of the power d (2 x −
1) = 10 x dx ; x dx = 110 d (2 x − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == 110 (cid:90) e x − d (2 x −
1) = 110 e x − + C. Result: ˙ I = 110 e x − + C, C ∈ R . ∗ Solution of Problem 1(c)guided by Irina Blazhievskais available on-line: https://youtu.be/kSn2UvdXWVs40 .(d) ˙ I = (cid:90) sin (1 + 3 x ) dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Applying the reduction power formula sin (1 + 3 x ) = 12 (cid:0) − cos(2 + 6 x ) (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == 12 (cid:90) (cid:0) − cos(2 + 6 x ) (cid:1) dx = 12 (cid:18)(cid:90) dx − (cid:90) cos(2 + 6 x ) dx (cid:19) == (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Substitution under the differential dx = 16 d (2 + 6 x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == 12 (cid:18) x − (cid:90) cos(2 + 6 x ) d (2 + 6 x ) (cid:19) + C = 12 x −
112 sin(2 + 6 x ) + C. Result: ˙ I = 12 x −
112 sin(2 + 6 x ) + C, C ∈ R . ˙ I = (cid:90) − x + 5 x − dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Checking the differential of the denominator d ( x −
4) = 2 xdx
Decomposing the numerator − x + 5 = − x ) + 5 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − (cid:90) x dxx − (cid:90) dxx − − (cid:90) d ( x − x − (cid:90) dxx − == − | x − | + 54 ln (cid:12)(cid:12)(cid:12)(cid:12) x − x + 2 (cid:12)(cid:12)(cid:12)(cid:12) + C = −
34 ln | x − | −
134 ln | x + 2 | + C. Result: ˙ I = −
34 ln | x − | −
134 ln | x + 2 | + C, C ∈ R . ˙ I = (cid:90) dx √ x − x dx = (cid:12)(cid:12)(cid:12)(cid:12) Completing the square inside the radical x − x = 1 − (3 x − (cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) dx (cid:112) − (3 x − = 13 (cid:90) d (3 x − (cid:112) − (3 x − = 13 arcsin(3 x −
1) + C. Result: ˙ I = 13 arcsin(3 x −
1) +
C, C ∈ R . ∗ Solution of Problem 2(b)guided by Irina Blazhievskais available on-line: https://youtu.be/KQuEzkh6AqQ41 .(c) ˙ I = (cid:90) − x − x + 6 x + 10 dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Checking the differential of the denominator: d ( x + 6 x + 10) = (2 x + 6) dx ; Decomposing the numerator − x − − (2 x + 6) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − (cid:90) x + 6 x + 6 x + 10 dx − (cid:90) dxx + 6 x + 10 = (cid:12)(cid:12)(cid:12)(cid:12) Completing the square: x + 6 x + 10 = ( x + 3) + 1 (cid:12)(cid:12)(cid:12)(cid:12) == − (cid:90) d ( x + 6 x + 10) x + 6 x + 10 − (cid:90) d ( x + 3)( x + 3) + 1 == − ln( x + 6 x + 10) − arctg( x + 3) + C. Result: ˙ I = − ln( x + 6 x + 10) − arctg( x + 3) + C, C ∈ R . ∗ Solution of Problem 2(c)guided by Ricard Riba isavailable on-line: https://youtu.be/RXJcMkNz8Zg ˙ I = (cid:90) x − √ x + 4 x − dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Checking the differential of the quadratic function: d ( x + 4 x −
5) = (2 x + 4) dx ; Descomposing the numerator (8 x −
5) = 4(2 x + 4) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ==4 (cid:90) x + 4 √ x + 4 x − dx − (cid:90) dx √ x + 4 x − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Completing the square: x + 4 x − x + 2) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ==4 (cid:90) d ( x + 4 x − √ x + 4 x − − (cid:90) d ( x + 2) (cid:112) ( x + 2) − ==8 (cid:112) x + 4 x − −
21 ln (cid:12)(cid:12)(cid:12) x + 2 + (cid:112) x + 4 x − (cid:12)(cid:12)(cid:12) + C. Result: ˙ I = 8 √ x + 4 x − −
21 ln (cid:12)(cid:12) x + 2 + √ x + 4 x − (cid:12)(cid:12) + C, C ∈ R . ˙ I = (cid:90) x + 2 √ x + 4 dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Performing the change of variable: t = √ x + 4; x = t − , dx = 2 tdt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) t −
4) + 2 t tdt = 2 (cid:90) (3 t − dt = 2( t − t ) + C = (cid:12)(cid:12)(cid:12)(cid:12) Undoing the change of variable: t = ( x + 4) / (cid:12)(cid:12)(cid:12)(cid:12) = 2( x + 4) / − x + 4) / + C. Result: ˙ I = 2( x + 4) / − x + 4) / + C, C ∈ R . ∗ Solution of Problem 3(a)guided by Irina Blazhievskais available on-line: https://youtu.be/ZOFSo2oDVzQ ˙ I = (cid:90) dxx √ x − x + 1 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Performing the change of variable: t = 1 x ; x = 1 t , dx = − t dt ; √ x − x + 1 = √ t − t + 3 t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) − t dt t · √ t − t + 3 t = − (cid:90) dt √ t − t + 3 = − (cid:90) d ( t − (cid:112) ( t − − − ln (cid:12)(cid:12)(cid:12) t − (cid:112) t − t + 3 (cid:12)(cid:12)(cid:12) + C = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Undoing the change of variable: t = 1 x ; √ t − t + 3 = √ x − x + 1 x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − ln (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − √ x − x + 1 x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + C. Result: ˙ I = − ln (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − √ x − x + 1 x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + C, C ∈ R . ˙ I = (cid:90) x − e x dx = (cid:90) (4 x − e − x dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Using integration by parts: (cid:82) udv = uv − (cid:82) vdu,u = 4 x − dv = e − x dx ; du = 8 xdx ; v = − e − x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − e − x (4 x − − (cid:90) (cid:18) − (cid:19) xe − x dx = − e − x (4 x −
3) + 4 (cid:90) xe − x dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Using integration by parts: (cid:82) udv = uv − (cid:82) vdu,u = x ; dv = e − x dx ; du = dx ; v = − e − x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − e − x (4 x −
3) + 4 (cid:18) − xe − x − (cid:90) (cid:18) − (cid:19) e − x dx (cid:19) == −
12 (4 x − e − x − xe − x − e − x + C = (cid:18) − x − x + 12 (cid:19) e − x + C. Result: ˙ I = (cid:0) − x − x + (cid:1) e − x + C, C ∈ R . ˙ I = (cid:90) x ln xdx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Using integration by parts: (cid:82) udv = uv − (cid:82) vdu,u = ln x ; dv = x dx ; du = dxx ; v = 13 x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 13 x ln x − (cid:90) x dxx == 13 x ln x − (cid:90) x dx = 13 x ln x − · x + C = 19 x (3 ln x −
1) + C. Result: ˙ I = 19 x (3 ln x −
1) +
C, C ∈ R . ∗ Solution of Problem 3(d)guided by Ricard Riba isavailable on-line: https://youtu.be/hDYt-m7ZCgM ˙ I = (cid:90) sin √ x + 1 dx = (cid:12)(cid:12)(cid:12)(cid:12) Performing the change of variable: t = √ x + 1; x = t − , dx = 2 tdt (cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) t sin tdt = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Using integration by parts: (cid:82) udv = uv − (cid:82) vdu,u = 2 t ; dv = sin tdt ; du = 2 dt ; v = − cos t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − t cos t − (cid:90) ( − cos t )2 dt == − t cos t + 2 sin t + C = (cid:12)(cid:12) Undoing the change of variable: t = √ x + 1 (cid:12)(cid:12) == − √ x + 1 cos √ x + 1 + 2 sin √ x + 1 + C. Result: ˙ I = − √ x + 1 cos √ x + 1 + 2 sin √ x + 1 + C, C ∈ R . .(f ) ˙ I = (cid:90) x sin x cos x dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Using integration by parts: (cid:82) udv = uv − (cid:82) vduu = x ; v = (cid:82) sin x cos x dx = − (cid:82) d (cos x )cos x = 12 cos x ; du = dx ; dv = sin x cos x dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == x x − (cid:90) dx cos x = x x −
12 tg x + C = 12 (cid:16) x cos x − tg x (cid:17) + C. Result: ˙ I = 12 (cid:16) x cos x − tg x (cid:17) + C, C ∈ R . ˙ I = (cid:90) − x + 2 x + 13 x + 2 x − x − dx Algorithm:
Step 1.
The function under the integral is a suitable polynomial fraction: f ( x ) = Q ( x ) P ( x ) . Step 2.
Finding the roots of the denominator P ( x ) = x + 2 x − x − . Possibleinteger roots may be ± ± .P (1) = 1 + 2 · − − ⇒ x = 1 is a root of P ( x ) ⇒ P ( x ) is divisible by ( x − . The decomposition P ( x ) = ( x − x + 3 x + 2) = ( x − x + 2)( x + 1) , impliesthat all roots are simple and real-valued. Step 3.
Applying the method of unknown coefficients to the suitable fraction: f ( x ) = − x + 2 x + 13( x − x + 2)( x + 1) = Ax − Bx + 1 + Cx + 2 == A ( x + 2)( x + 1) + B ( x − x + 2) + C ( x − x + 1)( x − x + 2)( x + 1) . Numerator’s equality: − x + 2 x + 13 = A ( x + 2)( x + 1) + B ( x − x + 2) + C ( x − x + 1) . x = 1 ⇒ − A (1 + 1)(1 + 2); 12 = 6 A ⇒ A = 2 x = − ⇒ − − B ( − − − − B ⇒ B = − x = − ⇒ − · − C ( − − − − C ⇒ C = − . The resulting decomposition is as follows: f ( x ) = − x + 2 x + 13( x − x + 2)( x + 1) = 2 x − − x + 1 − x + 2 . Step 4.
Substitution of the fully decomposed fraction under the integral: (cid:90) − x + 2 x + 13 x + 2 x − x − dx = (cid:90) (cid:18) x − − x + 1 − x + 2 (cid:19) dx == 2 ln | x − | − | x + 1 | − ln | x + 2 | + C. Result: ˙ I = 2 ln | x − | − | x + 1 | − ln | x + 2 | + C, C ∈ R . ˙ I = (cid:90) x − x + 56 x − x − x + 8 dx Algorithm:
Step 1.
The function under the integral is a non-suitable polynomial fraction: f ( x ) = Q ( x ) P ( x ) . Separation of integer part leads us to the following sum of 0-degree polynomialand a suitable polynomial fraction: f ( x ) = 3 + 6 x − x + 32 x − x − x + 8 . Step 2.
Finding the roots of the denominator P ( x ) = x − x − x + 8 . Possible integer roots may be ± , ± , ± , ± .P (2) = 2 − · − · ⇒ x = 8 is a root of P ( x ) ⇒ P ( x ) is divisible by ( x − . The decomposition P ( x ) = ( x − x −
4) = ( x − ( x + 2) , implies that P has two real-valued roots, a double root x = 2 and a simple root x = − . tep 3. Applying the method of unknown coefficients to the suitable fraction: x − x + 32 x − x − x + 8 = 6 x − x + 32( x − ( x + 2) = Ax − B ( x − + Cx + 2 == A ( x − x + 2) + B ( x + 2) + C ( x − ( x − ( x + 2) . Numerator’s equality: x − x + 32 = A ( x − x + 2) + B ( x + 2) + C ( x − == x ( A + C ) + x ( B − C ) + ( − A + 2 B + 4 C ) . Mix of equalities by zeros with equality of coefficients of some monomial: x = 2 ⇒ · − · B (2 + 2); 16 = 4 B ⇒ B = 4 x = − ⇒ · ( − + 32 = C ( − − ; 96 = 16 C ⇒ C = 6 x : 6 = A + C ; A = 6 − C = 6 − ⇒ A = 0 . The resulting decomposition is as follows: x − x + 32 x − x − x + 8 = 6 x − x + 32( x − ( x + 2) = 4( x − + 6 x + 2 . Step 4.
Substitution of the fully decomposed fraction under the integral: (cid:90) x − x + 56 x − x − x + 8 dx = (cid:90) (cid:18) x − + 6 x + 2 (cid:19) dx == 3 x − x − | x + 2 | + C. Result: ˙ I = 3 x − x − | x + 2 | + C, C ∈ R . ∗ Solution of Problem 4(b)guided by Irina Blazhievskais available on-line: https://youtu.be/X_300OO1i8A ˙ I = (cid:90) x − x − x + 4 x + 5)( x − dx Step 1.
The function under the integral is a suitable polynomial fraction: f ( x ) = Q ( x ) P ( x ) . Step 2.
Finding the roots of the denominator P ( x ) = ( x + 4 x + 5)( x − . The polynomial x + 4 x + 5 is indecomposable quadratic function since its dis-criminant is negative. The decomposition of P ( x ) implies that the unique real-valued root of P ( x ) is x = 1 . Step 3.
Applying the method of unknown coefficients to the suitable fraction: f ( x ) = x − x − x + 4 x + 5)( x −
1) = Ax + Bx + 4 x + 5 + Cx − Ax + B )( x −
1) + C ( x + 4 x + 5)( x + 4 x + 5)( x − . Numerator’s equality: x − x − Ax + B )( x −
1) + C ( x + 4 x + 5) == x ( A + C ) + x ( − A + B + 4 C ) + ( − B + 5 C ) . Mix of equalities by zeros with equality of coefficients of some monomial: x = 1 ⇒ − − C (1 + 4 + 5); −
10 = 10 C ⇒ C = − x : 1 = A + C ; A = 1 − C = 1 − ( −
1) = 2 ⇒ A = 2 x : − − B + 5 C ; B = 9 + 5 C = 9 + 5 · ( −
1) = 4 ⇒ B = 4 . The resulting decomposition is as follows: f ( x ) = x − x − x + 4 x + 5)( x −
1) = 2 x + 4 x + 4 x + 5 − x − . Step 4.
Substitution of the fully decomposed fraction under the integral: (cid:90) x − x − x + 4 x + 5)( x − dx = (cid:90) (cid:18) x + 4 x + 4 x + 5 − x − (cid:19) dx == (cid:90) d ( x + 4 x + 5) x + 4 x + 5 − (cid:90) dxx − x + 4 x + 5) − ln | x − | + C. Result: ˙ I = ln( x + 4 x + 5) − ln | x − | + C, C ∈ R . .(a) ˙ I = (cid:90) sin 10 x sin 3 xdx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Applying the decomposition ofa product of sinus functions: sin α sin β = 12 (cid:0) cos( α − β ) − cos( α + β ) (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == 12 (cid:90) (cid:0) cos(10 x − x ) − cos(10 x + 3 x ) (cid:1) dx = 12 (cid:90) (cid:0) cos 7 x − cos 13 x (cid:1) dx == 12 · sin 7 x − · sin 13 x
13 + C = 114 sin 7 x −
126 sin 13 x + C. Result: ˙ I = 114 sin 7 x −
126 sin 13 x + C, C ∈ R . ∗ Solution of Problem 5(a)guided by Irina Blazhievskais available on-line: https://youtu.be/-Zsc5t-YIOk ˙ I = (cid:90) (cid:114) sin x cos x dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Applying the integration of a product sin m x cos n x with m = 0 , n = − , decompose :cos − / x = cos − / x cos − x ;1cos x = 1 + tg x ; dx cos x = d (tg x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) (cid:114) sin x cos x dx cos x = (cid:90) tg / x x dx cos x = (cid:90) tg / x (cid:0) x (cid:1) d (tg x ) == (cid:90) (cid:0) tg / x + tg / x (cid:1) d (tg x ) = 34 tg / x + 310 tg / x + C. Result: ˙ I = 34 tg / x + 310 tg / x + C, C ∈ R . ˙ I = (cid:90) dx x + 3 sin x + 6 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) R (sin x, cos x ) = 14 cos x + 3 sin x + 6 has a general form.Applying the universal subtitution: t = tg x dx = 2 dt t ; sin x = 2 t t , cos x = 1 − t t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) dt t − t t + 3 2 t t + 6 = (cid:90) dt − t ) + 6 t + 6(1 + t ) = (cid:90) dtt + 3 t + 5 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Completing the square: t + 3 t + 5 == (cid:0) t + (cid:1) − (cid:0) (cid:1) + 5 = (cid:0) t + (cid:1) + (cid:16) √ (cid:17) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:90) d (cid:0) t + (cid:1)(cid:0) t + (cid:1) + (cid:16) √ (cid:17) == 1 √ arctg (cid:32) t + √ (cid:33) + C = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Undoingthe changeof variable: t = tg x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 √
11 arctg x √ + C. Result: ˙ I = 2 √
11 arctg x √ + C, C ∈ R . ∗ Solution of Problem 5(c)guided by Irina Blazhievskais available on-line: https://youtu.be/loxi3dwTmho ˙ I = (cid:90) dxx √ − x = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Performing the change of variable: x = 2 sin t ; dx = 2 cos tdt ; √ − x = 2 cos t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:90) tdt (2 sin t ) t == 14 (cid:90) dt sin t = −
14 ctg t + C = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Undoing the change of variable: ctg t = 2 cos t t = √ − x x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − √ − x x = − (cid:114) x − C. Result: ˙ I = − (cid:114) x − C, C ∈ R . ∗ Solution of Problem 6(a)guided by Ricard Riba isavailable on-line: https://youtu.be/ba4kyucyLVk50
Alternative solution 1 ˙ I = (cid:90) dxx √ − x = (cid:90) dxx (cid:114) x − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Substitution under the differential d (cid:18) x − (cid:19) = − dxx dxx = − d (cid:18) x − (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − (cid:90) (cid:18) x − (cid:19) − / d (cid:18) x − (cid:19) = − (cid:18) x − (cid:19) / + C = − (cid:114) x − C. • Alternative solution 2 ˙ I = (cid:90) dxx √ − x = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Performing the changeof variable: t = 1 xx = 1 t , dx = − t dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:90) − t dt t (cid:115) − (cid:18) t (cid:19) == − (cid:90) t √ t − dt = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d (4 t −
1) = 8 tdt ; dt = 18 d (4 t − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − (cid:90) d (4 t − √ t − − (cid:112) t − C = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Undoing the changeof variable: t = 1 x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − (cid:114) x − C. • Alternative solution 3 ˙ I = (cid:90) dxx √ − x = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Performing the change of variable: x = 2 th t, dx = 2 dt ch t ; √ − x = (cid:112) − t = 2ch t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:90) dt ch t (2 th t ) t == 14 (cid:90) ch tdt sh t = 14 (cid:90) d (sh t )sh t = −
14 1sh t + C = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Undoing the change ofvariable: sh t = 2 th t ch t x √ − x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − √ − x x + C = − (cid:114) x − C. .(b) ˙ I = (cid:90) (cid:114) x + 1 x − dx ( x − = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Performing the change of variable: t = (cid:114) x + 1 x − x + 1 x − t ; x = t + 1 t − t − x − t − dx = − t ( t − dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) t · − t ( t − (cid:18) t − (cid:19) dt = − (cid:90) t ( t − dt = − (cid:90) ( t − t ) dt == − (cid:18) t − t (cid:19) + C = − t + 316 t + C == (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Undoing the change of variable: t = (cid:16) x + 1 x − (cid:17) / (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − (cid:18) x + 1 x − (cid:19) / + 316 (cid:18) x + 1 x − (cid:19) / + C. Result: ˙ I = − (cid:18) x + 1 x − (cid:19) / + 316 (cid:18) x + 1 x − (cid:19) / + C, C ∈ R . ˙ I = (cid:90) ln(1 + x ) dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Using integration by parts: (cid:82) ba udv = uv | ba − (cid:82) ba vdu,u = ln(1 + x ); dv = dx ; du = 2 xdx x ; v = x. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == x ln(1 + x ) (cid:12)(cid:12)(cid:12) − (cid:90) x x dx = ln 2 − (cid:90) ( x + 1) −
11 + x dx == ln 2 − (cid:90) (cid:18) −
11 + x (cid:19) dx = ln 2 − (cid:104) x − arctg x (cid:105)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == ln 2 + ( − − (0 + 2 arctg 0) = ln 2 + π − . Result: ˙ I = ln 2 + π − . ∗ Solution of Problem 7(a)guided by Irina Blazhievskais available on-line: https://youtu.be/jfbI2G23U2M52 .(b) ˙ I = π/ (cid:90) sin x √ cos xdx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Applying the integration of a product sin m x cos n x with m = 3 , n = 0 decompose :sin x = sin x sin x ;sin x = 1 − cos x ; sin xdx = − d (cos x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == − π/ (cid:90) (1 − cos x ) cos / xd (cos x ) = π/ (cid:90) (cos / x − cos / x ) d (cos x ) == (cid:20)
413 cos / x −
45 cos / x (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π/ = 0 − (cid:20) − (cid:21) = 3265 . Result: ˙ I = 3265 . ∗ Solution of Problem 7(b)guided by Irina Blazhievskais available on-line: https://youtu.be/s4VH2LvXh7M ˙ I = √ (cid:90) x dx √ − x = √ (cid:90) x dx (cid:112) − ( x ) = (cid:12)(cid:12)(cid:12)(cid:12) Substitution under the differential x dx = d ( x ) (cid:12)(cid:12)(cid:12)(cid:12) == √ (cid:90) d ( x ) (cid:112) − ( x ) = arcsin x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) √ = arcsin √ − arcsin 12 = π − π π . Result: ˙ I = π . ˙ I = (cid:90) √ x √ x + √ x dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Performing the change of variable: t = √ xx = t , dx = 4 t dt ; x = 1 (cid:55)→ t = 1 ,x = 16 (cid:55)→ t = 2 x 1 16t 1 2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == (cid:90) t t + t t dt = 4 (cid:90) t + t t dt = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Separating the fraction’s integer part t + t t = t − t + 2 t − t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ==4 (cid:90) (cid:18) t − t + 2 t − t (cid:19) dt = (cid:20) t − t + 4 t − t + 8 ln | t | (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:18) −
323 + 16 −
16 + 8 ln 3 (cid:19) − (cid:18) −
43 + 4 − (cid:19) = 293 + 8 ln 32 . Result: ˙ I = 293 + 8 ln 32 . ∗ Solution of Problem 7(d)guided by Irina Blazhievskais available on-line: https://youtu.be/gCX8MgQrd7A
Find the area of the figure bounded by the curves: y = 2 x − x + 6; y = x − x. ∗ Solution of Problem 8(a)guided by Irina Blazhievskais available on-line: https://youtu.be/w4wx7Dd547wThe written version of solution is proposed below. Algorithm:
Step 1.
Building the picture. y = x - + = x - ( x ) Figure's region y = 2 x − x + 6 is a ∪ -shapedparabola with vertex (cid:0) , − (cid:1) and points of intersection withOX: x = −√ ≈ , and x = √ ≈ , .y = x − x is a ∪ -shapedparabola with vertex (cid:0) , − (cid:1) and points of intersection withOX: x = 0 and x = 3 . Step 2.
Finding the points of intersection between the curves. (cid:26) y = 2 x − x + 6 y = x − x ⇒ x − x + 6 = x − x ; x − x + 6 =0;( x − x −
1) =0;
Abscises of points: x = 1 , x = 6 . tep 3. Analytical description of the region in Cartesian coordinates: D = (cid:8) ≤ x ≤
6; 2 x − x + 6 ≤ y ≤ x − x (cid:9) . Step 4.
Applying the suitable formula to find the area. S ( D ) = b (cid:90) a (cid:0) y ( x ) − y ( x ) (cid:1) dx = (cid:90) (cid:0) x − x − (2 x − x + 6) (cid:1) dx == (cid:90) (7 x − x − dx = (cid:20) x − x − x (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == 72 (36 − −
13 (216 − − −
1) = 5 (cid:18) − − (cid:19) = 1256 . Result: S ( D ) = 1256 ( square units ) . Find the area of the figure bounded by the curves: x ( t ) = 4 cos t, y ( t ) = 4 sin t, y = 12 (cid:18) y ≥ (cid:19) . Algorithm:
Step 1.
Building the picture. x = t, y = ty = - - - - x ( t ) = 4 cos t, y ( t ) = 4 sin t is aparametric representation of an as-troid inscribed inside a circle of ra-dius 4. y = is the straight line passingthrough (cid:0) , (cid:1) and parallel to OX. Step 2.
Finding the points of inter-section between the curves. (cid:40) y = 4 sin ty = 12 ⇒ sin t = 12 ⇒ Parameters of points: t = π , t = 5 π . tep 3. Analytical description of the region in Parametric coordinates.The OY-symmetry of region D implies that its area is: S ( D ) = 2 S ( D + ) , where D + is the subregion contained in the first quadrant, D + = (cid:26) ≤ x ≤ t ; 12 ≤ y ≤ t, π ≤ t ≤ π (cid:27) . Step 4.
Applying the suitable formula to find the area. S ( D ) =2 S ( D + ) = 2 β (cid:90) α x ( t ) y (cid:48) ( t ) dt = (cid:12)(cid:12)(cid:12)(cid:12) x ( t ) = 4 cos t,y (cid:48) ( t ) = (4 sin t ) (cid:48) = 12 sin t cos t (cid:12)(cid:12)(cid:12)(cid:12) ==2 · · π/ (cid:90) π/ cos t sin tdt = 96 π/ (cid:90) π/ t · sin t dt ==12 π/ (cid:90) π/ (cid:18) − cos 4 t t sin t (cid:19) dt = 12 (cid:20) t −
18 sin 4 t + 16 sin t (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π/ π/ ==12 (cid:20) (cid:16) π − π (cid:17) − (cid:18) sin 2 π − sin 4 π (cid:19) + 16 (cid:16) sin π − sin π (cid:17)(cid:21) ==12 π √ − (cid:32) √ (cid:33) = 12 π π. Result: S ( D ) = 2 π ( square units ) . Find the area of the figure bounded by the curve: ρ = 4 sin 3 φ. Algorithm:
Step 1.
Building the picture. ρ = 4 sin 3 φ is a polar representation of a 3-petaled rose inscribed inside a circleof radius 4. Since this curve is constructed from a sin-function and negativeradius is not allowed, it has OY-symmetry and it is well-defined when π k ≤ φ ≤ π + π k, k ∈ { , , } . π implies that the rose’s k-petal is createdby rotating 0-petal in counterclockwise direction around the pole an angle π k,k ∈ { , } . Next we show the building of the -petal. For this, we set φ ∈ (cid:2) , π (cid:3) and consider the additional table. φ π π π π π π π π ρ = 4 sin 3 φ √ √ √ √ Step 2.
Analytical description of the region in Polar coordinates. - - - - Since the rose is formed by rota-tion of 0-petal around the pole, allpetals have the same metric charac-teristics.Then the 3 petals have the samearea. The multiplication of the -petal’s area by give us the area ofthe whole region D : S ( D ) = 3 S ( D ) , where D is the 0-petal’s region, D = (cid:110) ≤ φ ≤ π ≤ ρ ≤ φ (cid:111) Step 3.
Applying the suitable formula to find the area. S ( D ) =3 S ( D ) = 32 β (cid:90) α ρ ( φ ) dφ = 32 π/ (cid:90) (4 sin 3 φ ) dφ = 24 π/ (cid:90) sin φdφ ==12 π/ (cid:90) (1 − cos 6 φ ) dφ = 12 (cid:20) φ −
16 sin 6 φ (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π/ = 12 (cid:18) π −
16 sin 2 π (cid:19) = 4 π. Result: S ( D ) = 4 π ( square units ) . Find the arc length of the curve y = 13 ln(cos 3 x ) , ≤ x ≤ π . Step 1.
Building the picture. - - - - - - ( x ) Curve's arc y = ln(cos 3 x ) is acurve constructed from acos-function. Since the log-arithm is only defined forpositive values, this curveis only defined on regions (cid:2) − π + πk , π + πk (cid:3) , k ∈ Z . On each region it has a (cid:84) -like form with zeros-maximaat x = π k, and verticalasymptotes: x = ± π + πk , k ∈ Z . Step 2.
Analytical description of the arc segment in Cartesian coordinates:
Γ = (cid:26) y = 13 ln(cos 3 x ); 0 ≤ x ≤ π (cid:27) Note that (cid:2) , π (cid:3) belongs to the region of curve’s well-definiteness. Step 3.
Applying of the suitable formula to find the arc length. l (Γ) = b (cid:90) a (cid:112) y (cid:48) x ) dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y = 13 ln(cos 3 x ); y (cid:48) x = 13 · − x cos 3 x = − tg 3 x ;1 + ( y (cid:48) x ) = 1 + ( − tg 3 x ) = 1cos x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == π/ (cid:90) (cid:112) xdx = π/ (cid:90) | cos 3 x | dx = π/ (cid:90) x dx == π/ (cid:90) cos 3 x cos x dx = − π/ (cid:90) d (sin 3 x )sin x − dx = −
13 12 ln (cid:12)(cid:12)(cid:12)(cid:12) sin 3 x − x + 1 (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π/ == −
16 ln (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sin (cid:16) · π (cid:17) − (cid:16) · π (cid:17) + 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 16 ln (cid:12)(cid:12)(cid:12)(cid:12) −
10 + 1 (cid:12)(cid:12)(cid:12)(cid:12) = −
16 ln (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −
112 + 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 16 ln 3 . Result: l (Γ) = 16 ln 3 ( units ) . .(b) Find the arc length of the curve x ( t ) = 4( t − sin t ) , y ( t ) = 4(1 − cos t ) , t ∈ [0 , π ] . Algorithm:
Step 1.
Building the picture. ( t ) y ( t ) Curve's arc (cid:26) x ( t ) = 4( t − sin t ) y ( t ) = 4(1 − cos t ) is a parametric representa-tion of a cycloid with a hor-izontal base y = 0 . This curve is generated by the trajectory of a marked point in a circle of radius4 rolling on the "positive" side of the base. The parameter region ≤ t ≤ π creates one full wave of the curve, which intersects OX at points x = 0 and x = 8 π , has maxima at (4 π, and an axis of symmetry at x = 4 π . Step 2.
Analytical description of the arc segment in Parametric coordinates:
Γ = { x = 4( t − sin t ) , y = 4(1 − cos t ); 0 ≤ t ≤ π } . Step 3.
Applying the suitable formula to find the arc length. l (Γ) = β (cid:90) α (cid:112) ( x (cid:48) t ) + ( y (cid:48) t ) dt = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x (cid:48) t = (cid:16) t − sin t ) (cid:17) (cid:48) = 4(1 − cos t ); y (cid:48) t = (cid:16) − cos t ) (cid:17) (cid:48) = 4 sin t ;( x (cid:48) t ) + ( y (cid:48) t ) = 4 (cid:0) (1 − cos t ) + sin t (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ==4 π (cid:90) (cid:113) (1 − cos t ) + sin tdt = 4 π (cid:90) (cid:112) − cos t ) dt = 8 π (cid:90) (cid:12)(cid:12)(cid:12)(cid:12) sin t (cid:12)(cid:12)(cid:12)(cid:12) dt ==8 π (cid:90) sin t dt = −
16 cos t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π = − π − cos 0) = 32 . Result: l (Γ) = 32 ( units ) . .(c) Find the arc length of the curve ρ = 10 √ e φ , ≤ φ ≤ π. ∗ Solution of Problem 9(c)guided by Irina Blazhievskais available on-line: https://youtu.be/AJbHsr1eOesThe written version of solution is proposed below. Algorithm:
Step 1.
Building the picture. - - - - - - ρ = √ e φ is a polar represen-tation of a logarithmic spiral. Ithas 2 parameters; the initial ra-dius √ and the rate of spiral’sincreasing k = ( k = ρ (cid:48) φ ρ ).Since k > , the spiral rotatesaround the pole in counterclock-wise direction with increasing po-lar radius.The angle domain ≤ φ ≤ π generates one full turn with con-tinuous increase of radius from √ to √ e π . Step 2.
Analytical description of the arc segment in Polar coordinates:
Γ = (cid:26) ρ = 10 √ e φ ; 0 ≤ φ ≤ π (cid:27) . Step 3.
Applying the suitable formula to find the arc length. l (Γ) = β (cid:90) α (cid:113) ρ + ( ρ (cid:48) φ ) dφ = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ρ = √ e φ ; ρ (cid:48) φ = √ · e φ = √ e φ ; ρ + ( ρ (cid:48) φ ) = e φ + e φ = (cid:0) e φ (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == π (cid:90) (cid:113)(cid:0) e φ (cid:1) dφ = π (cid:90) e φ dφ = 10 e φ (cid:12)(cid:12)(cid:12) π = 10( e π − e ) = 10( e π − . Result: l (Γ) = 10 (cid:0) e π − (cid:1) (units).60 Find the surface area generated by rotating the curve y = 12 ch 2 x, − ≤ x ≤ , around the axis l = OX.
Algorithm:
Step 1.
Building the picture. - - ( x ) Rotating curve's arc y = ch 2 x is a catenary with pa-rameter a = , minima (0 , ) . This curve is an idealized hangingchain or cable sags under its ownweight when it is supported onlyat its ends.The surface of revolution of the catenarycurve around OX is a catenoid.
Note thatthe catenoid has a minimal surface area.
Step 2.
Analytical description of therotating arc segment (the generatrix) inCartesian coordinates:
Γ = (cid:26) y = 12 ch 2 x ; − ≤ x ≤ (cid:27) Step 3.
Applying the suitable formula tofind the surface area. S OX = 2 π b (cid:90) a y ( x ) (cid:112) y (cid:48) x ) dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y = 12 ch 2 x ; y (cid:48) x = 12 · x = sh 2 x y (cid:48) x ) = 1 + (sh 2 x ) = ch x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == 2 π (cid:90) −
12 ch 2 x (cid:113) x ) dx = π (cid:90) − ch x dx = π (cid:90) − x dx == π (cid:20) x + 14 sh 4 x (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − = π (cid:18) (cid:19) = π (cid:16) (cid:17) . Result: S OX = π (cid:16) π (cid:17) (square units).61 Find the surface area generated by rotating the curve x = 3 + cos t, y = 2 + sin t, around the axis l = OY. ∗ Solution of Problem 10(b)guided by Irina Blazhievskais available on-line: https://youtu.be/nXzXyHIw0w8The written version of solution is proposed below. Algorithm:
Step 1.
Building the picture. ( t ) y ( t ) Rotating curve's arc (cid:26) x ( t ) = 3 + cos ty ( t ) = 2 + sin t is a parametric representation ofa circle of radius 1 with center in (3 , : ( x − + ( y − = 1 . The empty intersection of thecurve with OY implies that thereare no restrictions for parameter’sdomain: ≤ t ≤ π. The sur-face of revolution with this rotat-ing curve is a ring-torus.
Step 2.
Analytical de-scription of the rotating arcsegment (the generatrix) inParametric coordinates:
Γ = x = 3 + cos t,y = 2 + sin t ;0 ≤ t ≤ π Step 3.
Applying the suitable formula to find the surface area. S OY =2 π β (cid:90) α x ( t ) (cid:112) ( x (cid:48) t ) + ( y (cid:48) t ) dt = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ( t ) = 3 + cos t ; y ( t ) = 2 + sin t ; x (cid:48) t = (3 + cos t ) (cid:48) = − sin t ; y (cid:48) t = (2 + sin t ) (cid:48) = cos t ;( x (cid:48) t ) + ( y (cid:48) t ) = ( − sin t ) + (cos t ) = 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = π π (cid:90) (3 + cos t ) (cid:112) ( − sin t ) + (cos t ) dt = 2 π π (cid:90) (3 + cos t ) dt ==2 π (cid:104) t + sin t (cid:105)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π = 2 π [3(2 π −
0) + (sin 2 π − sin 0)] = 12 π . Result: S OY = 12 π (square units). Find the area surface area generated by rotating the curve ρ = (cid:112) cos 2 φ around the axis l = oρ Algorithm:
Step 1.
Building the picture. - - - - ρ = √ cos 2 φ is a Polar representationof an ∞ -shaped lemniscate inscribedinside a circle of radius 1. This curveis known as Bernoulli’s lemniscate.
Since this curve is constructed from acos-function and negative radius is notallowed, it has OX-symmetry and it iswell-defined on φ ∈ [ − π , π ] (cid:83) [ π , π ] .OX-symmetry of the curve implies therestriction on polar angle domain forrotating part: φ ∈ [0 , π ] ∪ [ π ; π ] . The surface of revolutiongenerated by the lemniscateis "Hourglass"-shaped and isattached on the right.
Step 2.
Analytical description of the rotating arc segment (the generatrix) inPolar coordinates.The mirror symmetry of the surface with respect to YZ-plane implies that bothHourglass’ sides have the same metric characteristics.63he multiplication of right-side surface area S + oρ by 2 gives us the the surfacearea of the whole solid of revolution: S oρ = 2 S + oρ , where the generatrix of S + oρ islocated in the first quadrant: Γ + = (cid:110) ρ = (cid:112) cos 2 φ ; 0 ≤ φ ≤ π (cid:111) . Step 3.
Applying the suitable formula to find the surface area. S oρ =2 S + oρ = 2 · π β (cid:90) α ρ ( φ ) sin φ (cid:113) ( ρ ) + ( ρ (cid:48) φ ) dφ = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ρ = √ cos 2 φ ; ρ (cid:48) φ = − sin 2 φ √ cos 2 φ ;( ρ ) + ( ρ (cid:48) φ ) = 1cos 2 φ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ==4 π π/ (cid:90) (cid:112) cos 2 φ sin φ (cid:114) φ dφ = 4 π π/ (cid:90) sin φdφ = − π cos φ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π/ == − π (cid:16) cos (cid:18) π (cid:19) − cos 0 (cid:17) = − π (cid:16) √ − (cid:17) = 2 π (2 − √ . Result: S oρ = 2 π (cid:0) − √ (cid:1) (square units). Find the volume of the body formed by rotating region between the curves y = x + 2 x + 5 , y = 5 − x, around the axis l = OX.
Algorithm:
Step 1.
Building the picture. y = - xy = x + + - - - - ( x ) Rotating region y = x + 2 x + 5 is a ∪ -shaped parabola with ver-tex ( − , ; it has empty in-tersection with OX, but in-tersects OY at (0 , .y = 5 − x is a straight linepassing through the points (5 , , (0 , . tep 2. Finding the points of intersection between the curves. (cid:26) y = x + 2 x + 5 y = 5 − x ⇒ x + 2 x + 5 =5 − x ; x + 3 x = 0; x ( x + 3) = 0 Abscises of points: x = − , x = 0 . Step 3.
Analytical description of the rotating region in Cartesian coordinates: D = (cid:8) − ≤ x ≤ x + 2 x + 5 ≤ y ≤ − x (cid:9) The body generated by revolution of the region D around OX is attached below,on the right. Step 4.
Applying the suitable formula to find the volume of the body. V OX = V − V = π b (cid:90) a (cid:0) y ( x ) − y ( x ) (cid:1) dx == π (cid:90) − (cid:0) (5 − x ) − ( x + 2 x + 5) (cid:1) dx == π (cid:90) − ( − x − x − x − x ) dx == π (cid:20) − x − x − x − x (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − == π (cid:20) ( − − + 13 ( − · ( − (cid:21) == π (cid:20) − (cid:21) = 2525 π = 50 , π. Result: V OX = 50 , π (cubic units). Find the volume of the body formed by rotating the region between the curves x = 5 + 4 y − y , x = 5 , around the axis l = OY.
Algorithm:
Step 1.
Building the picture. 65 = 5 + 4 y − y = 9 − ( y − is a ⊃ -shaped horizontal parabola with vertex (9 , and points of intersection with OY : y = − and y = 5 .x = 5 is a straight line passing through (5 , and parallel to OY. x = + - y x = - - ( y ) y Rotating region Step 2.
Finding the points of inter-section between the curves. (cid:26) x = 5 + 4 y − y x = 5 ⇒ y − y =5; − y + 4 y = 0; y ( y −
4) =0
Ordinates of points: y = 0 , y = 4 . Step 3.
Analytical description ofthe rotating region in Cartesian co-ordinates: D = (cid:26) ≤ y ≤ ≤ x ≤ y − y (cid:27) . The bodygenerated bythe revolutionof the region D around OYis attached onthe left. Step 4.
Applying the suitable formula to find the volume of the body. V OY = V − V = π d (cid:90) c (cid:0) x ( y ) − x ( y ) (cid:1) dy = π (cid:90) (cid:0) (5 + 4 y − y ) − (cid:1) dy == π (cid:90) ( y − y + 6 y + 40 y ) dy = π (cid:20) y − y + 2 y + 20 y (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == π (cid:20)
15 4 − · + 2 · + 20 · (cid:21) = π · (cid:20) − (cid:21) = 7045 π = 140 , π. Result: V OY = 140 , π (cubic units).66 Find the volume of the body formed by rotating the curve ρ = 6(1 + cos φ ) , around the axis l = oρ. ∗ Solution of Problem 11(c)guided by Irina Blazhievskais available on-line: https://youtu.be/tr2pXX5GJJEThe written version of solution is proposed below. Algorithm:
Step 1.
Building the picture. ρ = 6(1 + cos φ ) is a polar representation of a cardioid with a parameter a = 6 .This curve has a heart’s shape with a "stalk" at the pole and oρ -symmetry, withangle domain ≤ φ ≤ π . Since it is constructed from a cos-function, it hasOX-symmetry, which implies the restriction on polar angle domain for rotatingpart: φ ∈ [0 , π ] . Step 2.
Analytical description of the rotating region in Polar coordinates: D = { ≤ φ ≤ π, ≤ ρ ≤ φ ) } - -
505 xy Rotating region
The body generated by revolution of the region D around the oρ -axis has ashape of an apple with the stalk at the origin; it is attached on the right.67 tep 3. Applying the suitable formula to find the volume of the body. V oρ = 2 π β (cid:90) α ρ ( φ ) sin φ dφ = 2 π π (cid:90) (6(1 + cos φ )) sin φ dφ == − π π (cid:90) (1 + cos φ ) d (1 + cos φ ) = − π (cid:2) (1 + cos φ ) (cid:3) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π == − π (cid:2) (1 + cos π ) − (1 + cos 0) (cid:3) = − π (cid:2) − (cid:3) = 476 π. Result: V oρ = 476 π (cubic units). ˙ I = ∞ (cid:90) − x − x + 4 dx = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( x ) = 2 x − x + 4 is a bounded continuosfunction for all x ∈ [ − , + ∞ ) ⇒ Improper integral of 1st kind (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == lim N →∞ N (cid:90) − x − x + 4 dx = lim N →∞ N (cid:90) − (cid:18) xx + 4 − x + 2 (cid:19) dx == lim N →∞ N (cid:90) − d ( x + 4) x + 4 − N (cid:90) − dxx + 2 = lim N →∞ (cid:20) ln( x + 4) −
12 arctg x (cid:21) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N − == lim N →∞ (cid:20) ln( N + 4) −
12 arctg (cid:18) N (cid:19) − ln(( − + 4) + 12 arctg (cid:18) − (cid:19)(cid:21) == lim N →∞ (cid:20) ln( N + 4) −
12 arctg (cid:18) N (cid:19) − ln 8 − π (cid:21) = + ∞ − π − ln 8 = + ∞ . Result:
The integral is divergent: ˙ I = + ∞ . ∗ Solution of Problem 12(a)guided by Irina Blazhievskais available on-line: https://youtu.be/r1GZaS5Hooo68 ˙ I = π/ (cid:90) e − tg x dx cos x = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( x ) = e − tg x cos x is a bounded continousfunction for all x ∈ [0 , π ); The endpoint x = π is a singular point of f ( x ) ⇒ Improper integral of 2nd kind (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == lim ε → + π/ − ε (cid:90) e − tg x dx cos x = lim ε → + π/ − ε (cid:90) e − tg x d (tg x ) = − lim ε → + e − tg x (cid:12)(cid:12)(cid:12) π/ − ε == − lim ε → + (cid:104) e − tg( π/ − ε ) − e tg 0 (cid:105) = 1 − lim ε → + e − ctg ε = 1 − . Result:
The integral is convergent: ˙ I = 1 . ∗ Solution of Problem 12(b)guided by Irina Blazhievskais available on-line: https://youtu.be/lbBT5mV_pc469 uthor’s Video-Lessons on Techniques of Integration (based on Sample task, page 39)
Indefinite Integration
Reduction to the tableof integrals: (cid:90) x (1 − x ) dx Guide: Ricard Riba Garcia https://youtu.be/x48CikKlF9cReduction to the tableof integrals: (cid:90) (cid:0) tg x (cid:1) dx Guide: Irina Blazhievska https://youtu.be/1MqmZbQ-3qMSubstitution under thedifferential: (cid:90) x e x − dx Guide: Irina Blazhievska https://youtu.be/kSn2UvdXWVsIntegration of fractionswith quadratic functions: (cid:90) dx √ x − x Guide: Irina Blazhievska https://youtu.be/KQuEzkh6AqQIntegration of fractionswith quadratic functions: (cid:90) ( − x − dxx + 6 x + 10 Guide: Ricard Riba Garcia https://youtu.be/RXJcMkNz8Zg70 ndefinite Integration
Change of variable: (cid:90) (3 x + 2) dx √ x + 4 Guide: Irina Blazhievska https://youtu.be/ZOFSo2oDVzQIntegration by parts: (cid:90) x ln xdx Guide: Ricard Riba Garcia https://youtu.be/hDYt-m7ZCgMIntegration of polynomialfractions: (cid:90) (3 x − x + 56) dxx − x − x + 8 Guide: Irina Blazhievska https://youtu.be/X_300OO1i8AIntegration of trigonometricfunctions (products): (cid:90) sin 10 x sin 3 xdx Guide: Irina Blazhievska https://youtu.be/-Zsc5t-YIOkIntegration of trigonometricrational function (fractions): (cid:90) dx x + 3 sin x + 6 Guide: Irina Blazhievska https://youtu.be/loxi3dwTmhoIntegration of fractionswith radicals: (cid:90) dxx √ − x Guide: Ricard Riba Garcia https://youtu.be/ba4kyucyLVk71 efinite Integration
Integration by parts: (cid:90) ln(1 + x ) dx Guide: Irina Blazhievska https://youtu.be/jfbI2G23U2MSubstitution under thedifferential: π/ (cid:90) sin x √ cos xdx Guide: Irina Blazhievska https://youtu.be/s4VH2LvXh7MChange of variable: (cid:90) (1 + √ x ) dx √ x + √ x Guide: Irina Blazhievska https://youtu.be/gCX8MgQrd7A
Geometric Applications of Definite Integrals • Cartesian coordinates:Area of the figure boundedby the curves: y = 2 x − x +6 , y = x − x Guide: Irina Blazhievska https://youtu.be/w4wx7Dd547w • Polar coordinates:Arc length of the curve: ρ ( φ ) = 10 √ e φ , φ ∈ [0 , π ] Guide: Irina Blazhievska https://youtu.be/AJbHsr1eOes72 eometric Applications of Definite Integrals • Parametric coordinates:Surface area generated byrotating the curve aroundOY-axis: x = 3 + cos t, y = 2 + sin t Guide: Irina Blazhievska https://youtu.be/nXzXyHIw0w8 • Cartesian coordinates:Volume of the bodygenerated by rotatingaround OX-axis the regionbounded by the curve: ρ ( φ ) = 6(1 + cos φ ) Guide: Irina Blazhievska https://youtu.be/tr2pXX5GJJE
Improper Integrals • Improper integralof 1st kind: ∞ (cid:90) − (2 x − dxx + 4 Guide: Irina Blazhievska https://youtu.be/r1GZaS5Hooo • Improper integralof 2nd kind: π/ (cid:90) e − tg x dx cos x Guide: Irina Blazhievska https://youtu.be/lbBT5mV_pc473 eferences [1] Берман Г.Н.
Сборник задач по курсу матемитического анализа : Уч. посо-бие. – 22-е изд. перераб. – СПб., Изд-во "Профессия 2005. – 432с.[2] Ефимов А.В., Демидович Б.П.
Сборник задач по матемитике для втузов.
Ч.1. Линейная алгебра и основы математического анализа: учебное пособиедля втузов./ В.А. Болгов, Б.П. Демидович, А.В. Ефимов, А.Ф. Каракулин,С.М. Коган, Е.Ф. Поршнева, А.С. Поспелов, Р.Я. Шостак. Пол ред. А.В.Ефимова и Б.П. Демидовича. – 2-е изд. – М.: Наука. Гл. ред. физ.-мат.лит., 1986. – 464с.[3] Письменный Д.Т.
Конспект лекций по высшей математике . Ч.1. – М.:Рольф, 2000. – 288с.[4] Fikhtengol’ts G.M.
The Fundamentals of Mathematical Analysis: InternationalSeries of Monographs in Pure and Applied Mathematics, Volume 2. – PergamonPress, 1965.– 540p.[5] Sagan H.
Advanced calculus of real-valued functions of a real variable and vector-valued functions of a vector variable . – Houghton Mifflin Company, 1974. –671p.[6] Spivak M.
Calculus .– 3rd edition (illustrated, reprint, revised). – CambridgeUniversity Press, 2006. – 670 p.
External links
Not only well-organized libraries of Math study, but also the interactive graphicsand infinity samples are proposed on the following cites: • • • • http://mathworld.wolfram.com • https://math.stackexchange.com ••