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reviewans - Math 116 Exam 1 Review Directions: Do the...

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Math 116 Exam 1 Review Directions: Do the following problems for extra credit by Friday, February 12. They are worth 5 points each, 30 points total. 1. Z ( x + 1) 2 ± 1 - 1 x ² dx = Z ( x 2 + 2 x + 1) ± 1 - 1 x ² dx = Z ± x 2 + x - 1 - 1 x ² dx = 1 3 x 3 + 1 2 x 2 - x - ln | x | + C 2. Z ± xe - x 2 + e x e x + 3 ² dx = Z xe - x 2 dx + Z e x e x + 3 dx = - 1 2 Z e u du + Z 1 v dv = - 1 2 e u + ln | v | + C = - 1 2 e - x 2 + ln | e x + 3 | + C In the first integral, the substitution is u = - x 2 , du = - 2 x dx . In the second integral, the substitution is v = e x + 3, dv = e x dx . 3. Z 4 0 e x x = Z 2 0 2 e u du = 2 e u | 2 0 = 2( e 2 - e 0 ) = 2 e 2 - 2 Use the substitution u = x , du = 1 2 x . 4. Given that Z 2 - 1 f ( x ) dx = 2 and Z 2 0 f ( x ) dx = 3, find Z 0 - 1 f ( x ) dx and Z 2 - 1 3 f ( x ) dx . Solution: Z 0 - 1 f ( x ) dx = Z 2 - 1 f ( x ) dx - Z 2 0 f ( x ) dx = 2 - 3 = - 1 Z 2 - 1 3 f ( x ) dx = 3 Z 2 - 1 f ( x ) dx = 3(2) = 6 5. Find the area between f ( x ) = 2 x and g ( x ) = x x + 1. Solution: Solve
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This note was uploaded on 04/06/2010 for the course MATH 116 taught by Professor Scholle,minho during the Spring '08 term at Kansas.

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reviewans - Math 116 Exam 1 Review Directions: Do the...

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