Thomas' Calculus: Early Transcendentals

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FINAL MAT104 SPRING 2006 SOLUTIONS Answers: (1) ln | x 2 + 1 + x | - x x 2 +1 + C (2) - ln(1+( x +1) 2 ) x +1 + 2 tan - 1 ( x + 1) + C (3) It converges. (4) a) It converges. b) It converges. (5) It converges for - 1 x < 1. (6) -5 (7) 2 49 / 2 e i 17 π 12 (8) y ( x ) = 1 2 e tan x + Ce - tan x (9) y ( x ) = C 1 e x + C 2 e - 2 x + 1 10 e 3 x (10) V = π 2 3 + π ( 3 - 1) (11) L = 38 Solutions: (1) Evaluate Z x 2 (1 + x 2 ) 3 / 2 dx. We use the trigonometric substitution x = tan θ , dx = sec 2 θdθ Z x 2 (1 + x 2 ) 3 / 2 dx = Z tan 2 θ sec 2 θ sec 3 θ = Z tan 2 θ sec θ = Z sec 2 θ - 1 sec θ = Z sec θ dθ - Z cos θ dθ = ln | sec θ + tan θ | - sin θ + C We have sin θ = x x 2 +1 and cos θ = 1 x 2 +1 . Rewriting the above result in terms of x we get Z x 2 (1 + x 2 ) 3 / 2 dx = ln | p x 2 + 1 + x | - x x 2 + 1 + C. (2) Evaluate Z ln( x 2 + 2 x + 2) ( x + 1) 2 dx. 1
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2 FINAL MAT104 SPRING 2006 SOLUTIONS We first make the change of variable y = x + 1, dy = dx to get Z ln( x 2 + 2 x + 2) ( x + 1) 2 dx = Z ln( y 2 + 1 y 2 dy. We then integrate by parts with u = ln(1 + y 2 ), du = 2 y 1+ y 2 dy and dv = 1 y 2 dy , v = - 1 y Z ln( y 2 + 1) y 2 dy = - ln(1 + y 2 ) y + Z 2 1 + y 2 dy = - ln(1 + y 2 ) y + 2 tan - 1 y + C. Rewriting in terms of the variable x Z ln( x 2 + 2 x + 2) ( x + 1) 2 dx = - ln(1 + ( x + 1) 2 ) x + 1 + 2 tan - 1 ( x + 1) + C (3) Does R 2 ln( e x - 2) x 3 +1 dx converge or diverge ? The only trouble spot is at . We have ln( e x - 2) x at x → ∞ as lim x →∞ ln( e x - 2) x = 1 using L’Hospital rule. Also x 3 + 1 x 3 at infinity. Therefore the convergence or divergence of the integral is the same as for the following integral Z 2 x x 3 dx = Z 2 1 x 2 dx. By the p -test, this integral converges. Hence the above integral converges.
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