108Fall12Ex1VersionASolutions

5 pts set up but do not evaluate or simplify the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 08/22/2013 for the course MTHS 1080 taught by Professor Briggs during the Spring '13 term at Clemson.

Ask a homework question - tutors are online