m129SGFa10Sols

# m129SGFa10Sols - MATH 129 FINAL EXAM REVIEW PACKET ANSWERS...

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MATH 129 FINAL EXAM REVIEW PACKET ANSWERS (Revised Fall 2010) 1. 3 0 10 sin 30 150 3 t dt π  +=   people. 2. 2 1 7 (5 2 ) 2 f x dx −= . Let 52 ux = and change the endpoints. 3. 32 12 2 ( 1) 2( 3 1 t dt t t c t = + −+ + + . By the method of substitution with 1 ut = + . You can also use integration by parts with = and ( vt ′ = + . The result is equivalent, just written in a different form. 4 2 ( ( 3 1 t dt t t t c t = +−++ + . 4. a) ( ) ( ) 2 2 2 ln 1 1 ln 1 2arctan z dz z z c zz + = ++ + . Let 2 ln( uz = and 2 1 v z ′ = . b) 2 22 4 11 arcsin( ) arcsin( ) 1 x x dx x x x c = + . First make a substitution with 2 wx = . Then let arcsin( ) uw = and 1 v ′ = . c) 1 0 () 3 x g x dx ′′ ⋅= . Let = and vg = for the first integration by parts. 5. a) 2 cos (3 2) cos(3 2)sin(3 2) (3 2) 66 dc θθ θ + = + + . Let u = before using table formula # 18. If you use another approach, your answer will look different. b) ( ) 2 21 ln 2 3 ln 2 3 49 6 dt t t c t = −− + + . Let 2 = and factor the denominator before using table formula # 26. If you use another approach, your answer will look different. c) 2 2 ln ( 4) 8 15 8 15 dy y yy c = ++ + + + . Complete the square before using table formula # 29. 6. 3 2 3 3 51 1 3 ln ln 1 2arctan( ) 2 dy y y y y c +− = + + + . First do long division. Then use partial fractions 2 1 A By C + + + .

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7. 2 32 2 1 (5 ) 5 5 dx x c x x = + Let 5sin( ) x θ = . 8. 22 () v kR r = . 2 0 12 03 R k R r dr kR R −= . 9. 0 7 0 1 ( ) 0 7 7 x E X xf x dx xf x dx xf x dx x e dx ∞∞ −∞ = = += + = ∫∫∫ . Use integration by parts or the table of integrals. 10. a) 2 1 1 16 9 16 0 11 0.75459794 t e dt e e −− ≈+≈ b) Not clear because 2 t ft e = changes concavity on the given interval. When () ft is concave down, the midpoint rule provides an overestimate.
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## This note was uploaded on 02/04/2011 for the course MATH 9 taught by Professor Dwang during the Spring '08 term at Arizona.

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m129SGFa10Sols - MATH 129 FINAL EXAM REVIEW PACKET ANSWERS...

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