Solutions to Tutorial Sheet 12 Topics: Integration by Substitution + Integration by the Method of Partial Fractions + Applications to Geometry

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1 The University of Sydney School of Mathematics and Statistics Solutions to Tutorial Sheet Topics: Integration by Substitution + Integration by the Method of Partial Fractions + Applications to Geometry MATH: Introduction to Calculus Semester, Web Page: Lecturer: Clio Cresswell. Evaluate the following integrals using the method of integration by substitution: (a e x (b xe x (c x +x (d (lnx x (e 6cos 5 θsinθdθ (a Let u = x, so du = and = du. Thus e x = e u du = eu +C = ex +C. (b Let u = x, so du = x and = x du. Thus xe x = xe u x du = e u du = eu +C = ex +C. (c Let u = +x, so du = x and = x du. Thus x = x +x u du = x u du = lnu+c = ln(+x +C. (d Let u = lnx, so du = x and = xdu. Thus (lnx x = u x xdu = u du = u (e Let u = cosθ, so du dθ = sinθ and dθ = sinθ du. +C = (lnx +C. Thus 6cos 5 θsinθdθ = 6 u 5 sinθ sinθ du = 6 u 5 du = u 6 +C = cos 6 θ+c.. Evaluate the following indefinite integrals: (a cosθ(sinθ +π 6 dθ (b costsintdt (c xsin(x (d x(x+ 7 (e x e x +x +e x +x (a Let u = sinθ +π, so du dθ = cosθ and so dθ = cosθdu. Thus cosθ(sinθ+π 6 dθ = cosθu 6 cosθ du = u 6 du = u7 7 +C = (sinθ+π7 7 +C. Copyright c The University of Sydney

2 (b Let u = cost, so du dt = sint and so dt = sintdu. Thus costsintdt = usint = u du = 6 u +C sint du = 6 (cost +C. (c Let u = x, so du = x and so = xdu. Thus xsin(x = xsinu sinudu = x du = cosu+c = cos(x +C. (d Let u = x+, so du = and so = du. Note also that x = u. Thus x(x+ 7 = x u 7 du = (u u 7 du = u 8 u 7 du = u9 9 u8 8 +C = (x+9 9 (x+8 8 +C. (e Let u = x +x, so du = x + and so = du. Thus x + x e x +x +e x +x = (x +e x +x. Evaluate the following definite integrals: (a x+ (b x x x+ (c x x+ (d sinθecosθ dθ = (x +e u x + du = e u du = e u du = e u +C = e x +x +C. (a Let u = x+, so du = and so = du. Note that when x =, u = and when x =, u = 6. Thus 6 x+ = udu [ 6 = ] u = 6 8 =.

3 (b Let u = x x+, so du = x and so = and when x =, u =. Thus x x x x+ = x = du. Note that when x =, u = u x du u du = [lnu] = ln ln. (c Let u = x+, so du = and so = du. Note that when x =, u = and when x =, u =. Note also that x = u. Thus x x+ = (u udu = u5 u +u du [ ] = frac7u 7 5 u5 + u = 6.. (d Let u = cosθ, so du dθ = sinθ and so dθ = sinθdu. Note that when θ = π, u = and when θ =, u =. Thus sinθecosθ dθ = sinθe u sinθ du = eu du = [e u ] = e e.. Find the area under the graph of y = cos ( θ sin ( θ between θ = and θ = π. We evaluate the following integral to get our area: ( ( θ θ cos sin dθ. We use integration by substitution; let u = sin ( θ, so du dθ = cos( θ Note that when θ =, u = and when θ = π, u =. Thus cos( θ sin ( θ dθ = cos( θ = = =. u du [ ] u u cos( θ du and so dθ = cos( θ du. 5. Explain why and π cosθdθ = cosθdθ =, π sinθdθ =. cos θ is an even function. The reflective symmetry about the y-axis means that the area under the curve between π and π is double the area under the curve between and π. sinθ is an odd function. The rotational symmetry about the origin means that the area under the curve between and π is the same as the area above the curve between π and, so these areas cancel each other out when integrating over the whole inverval.

4 6. Evaluate the indefinite integral x (x using two methods: (a Expand the polynomial inside the integral and then evaluate, (b Use integration by substitution. (c Do your answers agree? Why or why not? (a x (x = x 5 x = x6 x +C. (b Let u = x, so du = x and so = x du. Thus x (x = x u du x = udu = u +c = (x +c = x6 x + +c. (c The answers do agree. As C and c are arbitrary constants that could represent any value, setting C = +c gives the same answers in parts (a and (b. 7. Find the total area bound by the following curves: (a y = x and y = x (b y = 9 and y = x (c y = x +x x and the x-axis (d y = x x, x =, x = and the x-axis (a The two curves intersect when x = x, which occurs at x = and x =. On the interval [,], x x. Hence we evaluate [ ] x x = x x = 6. (b The two curves intersect when 9 = x, which occurs at x = and x =. On the interval [,], 9 x. Hence we evaluate [ ] 9 x = 9x x = 6. (c The curve y = x +x x = x(x (x+ intersects the x-axis at x =, x = and x =. On the interval [,], y and on the interval [,], y. Hence we evaluate Area = x +x x [ ] x +x x = x [ ] + x x x + x x = ( ( 8 (( + = 7. (d We evaluate x x using integration by substitution. let u = x, so du = x and so = xdu. When x =, u = and when x =, u = 8. Thus x x = 8 x u = = x du u du [ ] 8 u

5 8. Evaluate the following integrals using the method of integration by partial fractions: (a x x 5x (b x+ x 9 (c x+ x x 6 (d x x+6 (x (x+ (e x x+6 (x (x + (f x + x +x x (g 5 x (h x x +6x 7 x (a x 5x = x x(x 5 = A x + B for some A, B. This gives x = A(x 5 + Bx x 5 and hence A = /5 and B = 7/5. Therefore x x 5x = /5 x + 7/5 x 5 = 5 ln x 7 5 ln x 5 +C. x+ (b x 9 = x+ (x (x+ = A x + B for some A, B. This gives x+ = A(x++ x+ B(x and hence A = / and B = /. Therefore x+ x 9 = / x + / x+ = ln x + ln x+ +C. x+ (c x x 6 = x+ (x (x+ = A x + B for some A, B. This gives x + = x+ A(x++B(x and hence A = and B =. Therefore x+ x x 6 = x = ln x ln x+ +C. x+ x x+6 (d (x (x+ = A x + B (x + C x+ for some A, B, C. This gives x x+6 = A(x (x ++B(x+ +C(x and hence A = 9/, B = 9/ and C = /. Therefore x x+6 (x (x+ = 9/ x + 9/ (x + / 9 = x+ ln x 9 (x + ln x+ +C. x x+6 (e (x (x + = A x + Bx+C x + for some A, B, C. This gives x x + 6 = A(x + + (Bx + C(x and hence A = 6/5, B = /5 and C = 6/5. Therefore x x+6 (x (x + = 6/5 x + /5x 6/5 x = 6/5 + x x 5x + 6 5x + = 6 5 ln x ln x + 6 tan ( x +C. x + (f x +x x = x + x(x +x = x + x(x+(x = A x + B x+ + C for some A, x B, C. This gives x + = A(x+(x +Bx(x +Cx(x+ and hence A =, B = / and C = 5/. Therefore x + x +x x = x + / x+ + 5/ x = ln x + ln x ln x +C. 5

6 (g (h x = (x (x+ = A x + B for some A, B. Thisgives = A(x++B(x x+ and hence A = / and B = /. Therefore 5 [ ln x ln x+ ] 5 = ln ln7+ ln5. x x +6x 7 = x (x (x+7 = A x + B x+7 x = 5 / x / x+ = for some A, B. This gives x = A(x B(x and hence A = / and B = /. Therefore / x + / x+7 = [ ln x + ln x+7 ] = ln+ ln ln9. x x +6x 7 = 9. For each of the following functions and intervals, find the volume of the solid obtained by rotating the total region bound by the graph and the x-axis on the given interval about the x-axis. (a y = x, x (b y = x, x (c y = e x, x (d y = x, x (e y = x, x (a We evaluate π x = π [ ] x = 8π. (b We evaluate π x = π x = π (c We evaluate π (ex = π ex = π [ (d We evaluate π x = π (e Weevaluateπ 6π 5. [ ] x [ ] 5 x5 = π 5 ex] = π ( ( e e 8. = 6π. ( x = π x x + = π [ 5 x5 x +x ] = π( =. The rate at which the human body eliminates an experimental drug is modelled by the equation E(t = te t, where t is measured in days and E is measured in mg. If the initial dose is 5mg, how much of the drug remains after two and a half days? The amound of the drug in the body at any time x measured in days since the initial dose is A(x = 5 x te t dt. First we compute the integral.5 te t dt. Let u = t, so du dt = t and dt = tdu. Now when t =.5, u = 6.5, and when t =, u =. Hence.5 te t dt = 6.5 te u t du = eu du = 5 [ e e 6.5] = 5 5e 6.5. Thus A(.5 = 5 ( 5 5e 6.5 = 5e 6.5.mg. 6