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Unformatted text preview: lOMoARcPSD|5004096 Test 1 7 February, questions and answers Integral Calculus (University of Windsor) StuDocu is not sponsored or endorsed by any college or university Downloaded by Yarvnis Blogs ([email protected]) lOMoARcPSD|5004096 1 DEPARTMENT OF MATHEMATICS AND STATISTICS MATH 1730-3 Test 1 February 7, 2019 1. (a) (4 marks) Write out the form of the partial fraction decomposition of the function below. Do not calculate the coefficients A, B, C, etc. 5x3 + 4x − 8 (3x + 2)3 (x2 + 7) f (x) = Solution. f (x) = B C Dx + E A + + + 2 2 3 3x + 2 (3x + 2) (3x + 2) x +7 (b) (2 marks) Find the derivative g ′ (x) of the function g(x) = Z x3 (2t + 5)7 ln(t + 6) dt. 2 Solution. By the FTOC part 1 and the Chain Rule, we have g ′ (x) = (2x3 + 5)7 ln(x3 + 6) · (x3 )′ = (2x3 + 5)7 ln(x3 + 6) · 3x2 . (c) (2 marks) Evaluate Z 4 √ 7 + x8 sin7 x dx. −4 Solution. Let f (x) = So we have √ 7 + x8 sin7 x. Then f (−x) = −f (x), so f is an odd function. Z 4√ 7 + x8 sin7 x dx = 0. −4 Downloaded by Yarvnis Blogs ([email protected]) lOMoARcPSD|5004096 2 2. Evaluate the following integrals. (a) (5 marks) Z (3 ln x + 5)6 dx x Solution. Let u = 3 ln x + 5. Then du = x3 dx. µ ¶ Z Z Z (3 ln x + 5)6 1 1 1 1 6 u6 du = u7 + C = (3 ln x + 5)7 + C. dx = u du = x 3 3 21 21 (b) (6 marks) Z 1 xe5x dx 0 Solution. Let u = x, dv = e5x dx. Then du = dx, v = 51 e5x . Using integration by parts, Z Z 1 1 5 1 5x ¯¯1 1 1 5x 1 5x ¯¯1 1 4 5x e dx = e − e ¯ = e5 + . xe dx = xe ¯ − 5 5 0 5 25 25 25 0 0 0 (c) (6 marks) Z sin4 x cos3 x dx Solution. Let u = sin x. Then du = cos x dx. Z Z Z Z 4 3 4 2 4 2 sin x cos x dx = sin x (1 − sin x) cos x dx = u (1 − u ) du = (u4 − u6 ) du = 1 1 u5 u7 − + C = sin5 x − sin7 x + C. 5 7 5 7 Downloaded by Yarvnis Blogs ([email protected]) lOMoARcPSD|5004096 3 3. Evaluate the following integrals. (a) (6 marks) Z Solution. Let x−7 dx (2x + 1)(x − 2) x−7 A B = + . (2x + 1)(x − 2) 2x + 1 x − 2 Then x − 7 = A(x − 2) + B(2x + 1). Setting x = − 21 gives A = 3. Setting x = 2 gives B = −1. Z ³ Z 1 ´ 3 3 x−7 dx = − dx = ln |2x + 1| − ln |x − 2| + C. (2x + 1)(x − 2) 2x + 1 x − 2 2 (b) (6 marks) Z √ x2 dx. 9 − x2 Solution. Let x = 3 sin t, −π/2 < t < π/2. Then dx = 3 cos t dt, and p √ 9 − x2 = 9 − 9 sin2 t = 3 cos t. So Z Z Z Z 1 − cos 2t x2 (9 sin2 t)3 cos t 2 √ dt = 9 sin t dt = 9 dt dx = 3 cos t 2 9 − x2 ¢ ¢ 9¡ 1 9¡ = t − sin 2t + C = t − sin t cos t + C 2 2 2 √ 9 ³ −1 x x 9 − x2 ´ 9 x x√ 9 − x2 + C. = sin + C = sin−1 − − 2 3 3 3 2 3 2 (By using a right triangle or an identity we have cos t = Downloaded by Yarvnis Blogs ([email protected]) √ 9−x2 .) 3 lOMoARcPSD|5004096 4 4. (6 marks) Evaluate the improper integral Z ∞ 1 ln x dx. x4 Solution. Z ∞ ln x dx = lim t→∞ x4 1 Let u = ln x, dv = x Z 1 t x −4 −4 Z t 1 dx. Then du = ln x dx = lim t→∞ x4 1 x dx, v = Z t x−4 ln x dx. 1 − 13 x−3 . ¶ µ ¯t Z t 1 1 −3 1 ln t 1 1 ¯ −4 ln x dx = − x ln x¯ + x dx = − · 3 − −1 . 3 3 t 9 t3 1 1 3 By L’Hostital’s Rule we have lim t→∞ So Z ∞ 1 ln t 1/t 1 = lim 2 = lim 3 = 0. 3 t→∞ t→∞ t 3t 3t µ ¶¶ µ 1 ln t 1 1 1 ln x − dx = lim · − − 1 = . t→∞ x4 3 t3 9 t3 9 5. (7 mark) Evaluate the integral I= Solution. Let u = √ Z 1 √ dx 4 x−5 +x x − 5. Then u2 = x − 5, x = u2 + 5, dx = 2udu. Z Z 2u 2u I= du = du 4u + u2 + 5 (u + 2)2 + 1 Let t = u + 2. Then u = t − 2, du = dt. ¶ Z Z µ 2(t − 2) 4 2t I = dt = − dt t2 + 1 t2 + 1 t2 + 1 = ln(t2 + 1) − 4 tan−1 t + C √ ¢ ¡√ = ln ( x − 5 + 2)2 + 1 − 4 tan−1 ( x − 5 + 2) + C Downloaded by Yarvnis Blogs ([email protected]) ...
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