Cal 2 test 1 solutions - Math 1592 Calculus II- Sample Test...

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Unformatted text preview: Math 1592 Calculus II- Sample Test 1 Solutions 1. Z tan- 1 x dx Integration by parts. If u = tan- 1 x , v = x , du = dx 1 + x 2 , dv = dx , then Z tan- 1 x dx = x tan- 1 x- Z x 1 + x 2 dx and if u = 1 + x 2 , then du = 2 xdx so Z tan- 1 x dx = x tan- 1 x- 1 2 Z du u dx , = x tan- 1 x- 1 2 ln | u | + c , = x tan- 1 x- 1 2 ln | 1 + x 2 | + c . 2. Z ln x x 2 dx Integration by parts. If u = ln x , v =- 1 x , du = dx x , dv = dx x 2 , then Z ln x x 2 dx =- ln x x + Z dx x 2 =- ln x x- 1 x + c . 3. Z 1 xe x dx Integration by parts. If u = x , v = e x , du = dx , dv = e x dx , then Z 1 xe x dx = xe x | 1- Z 1 e x dx = xe x | 1- e x | 1 = ( e- )- ( e- 1 ) = 1. 1 4. Z sin- 1 x dx Integration by parts. If u = sin- 1 x , v = x , du = dx 1- x 2 , dv = dx , then Z sin- 1 x dx = x sin- 1 x- Z x 1- x 2 dx and if u = 1- x 2 , then du =- 2 xdx so Z sin- 1 x dx = x sin- 1 x + 1 2 Z du u dx , = x sin- 1 x + u + c , = x sin- 1 x + p 1- x 2 + c . 5. Z 1 dx 1- x 2 This integral is improper lim b 1 Z b dx 1- x 2 = lim b 1 sin- 1 x fl fl fl b = lim b 1 sin- 1 b = 2 . 6. Z 5 1 ln x dx Integration by parts. If u = ln x , v = x , du = dx x , dv = dx , then Z 5 1 ln x dx = x ln x | 5 1- Z 5 1 dx = x ln x | 5 1- x | 5 1 = 5 ln 5- 4. 7. Z sec 3 x dx Integration by parts. If u = sec x , v = tan x , du = sec x tan x dx , dv = sec 2 x dx , then Z sec 3 x dx = sec x tan x- Z sec x tan 2 x dx = sec x tan x- Z sec x ( sec 2 x- 1 ) dx = sec x tan x- Z sec 3 x dx + Z sec x dx 2 so 2 Z sec 3 x dx = sec x tan x + ln | sec x + tan x | , so Z sec 3 x dx = 1 2 sec x tan x + 1 2 ln | sec x + tan x | + c . 8. Z 3- 3 dx 4- x 2 dx This integral is improper because the integrand is undefined at x =- 2 and x = 2. So we consider Z 3- 3 dx 4- x 2 dx = Z- 2- 3 dx 4- x 2 + Z 2- 2 dx 4- x 2 + Z 3 2 dx 4- x 2 and will show that one of the integrals (the last one) diverges. Z 3 2 dx 4- x 2 = lim a 2 Z 3 a dx 4- x 2 dx = 1 4 lim a 2 Z 3 a 1 2 + x + 1 2- x dx = 1 4 lim a 2 ( ln | 2 + x | - ln | 2- x | ) | 3 a = 1 4 lim a 2 ( ln 5- ln | 2 + a | + ln | 2- a | ) =- ....
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This homework help was uploaded on 04/10/2008 for the course MATH 1592 taught by Professor Liu during the Spring '08 term at University of Central Arkansas.

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Cal 2 test 1 solutions - Math 1592 Calculus II- Sample Test...

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