108Fall12Ex1VersionASolutions

# 5 pts set up but do not evaluate or simplify the

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Unformatted text preview: ∫ eu du = − eu = − e1/ 2 − e ∫ x2 1 1 1 ( ) 2 e1/ x dx = e − e1/ 2 2 1x ⇒∫ x 2. (7 pts) ∫ dx 1 + x Let u = 1 + x. Then du = dx and x = u − 1. ∫ x 1+ x ⇒∫ dx = ∫ x 1+ x u −1 2 du = ∫ u1/ 2 − u −1/ 2 du = u 3/ 2 − 2u1/ 2 + C 1/ 2 3 u dx = ( ) 3/ 2 1/ 2 2 (1 + x ) − 2 (1 + x ) + C 3 MthSc 108 Test 1-Version A 2− x 3. (7 pts) ∫ 1 − 9x2 Fall 2012 dx ∫ 2− x 1− 9x 2 dx = ∫ 2 1− 9x 2 dx − ∫ x dx 1− 9x2 1 du = dx. 3 2 2 du 2 2 Therefore, ∫ dx = ∫ = sin −1 u + C = sin −1 ( 3x ) + C . 2 2 3 1− u 3 3 1− 9x 1 2 For the second integral, let w = 1 − 9 x . Then dw = −18 xdx , and − dw = xdx. 18 x 1 dw 1 1 Therefore, ∫ dx = − ∫ =− 2 w1/ 2 + C = − 1 − 9 x 2 + C . 18 18 9 w 1− 9x2 For the first integral, let u = 3x. Then du = 3dx , and ( ⇒∫ ) 2− x 2 1 dx = sin −1 ( 3x ) + 1− 9x2 + C 3 9 1− 9x 2 π /4 sec 2 x dx 4. (7 pts) ∫ tan x π /6 Let u = tan x. Then du = sec 2 xdx. π 1 π If x = then u = . If x = then u = 1. 6 4 3 π /4 1 1 sec 2 x du ∫/6 tan x dx = ∫ u = ln u 1/ π 1/ 3 ⇒ π /4 sec 2 x 1 ∫/6 tan x dx = 2 ln 3 π 3 = ln 1 − ln 1 3 ( = 0 − ln 1 − ln 3 ) 7 MthSc 108 Test 1-Version A Fall 2012 5. Let R be the region in the first...
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