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20b_03f_1s

# 20b_03f_1s - Math 20B First Midterm Solutions 1(a Integrate...

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Math 20B First Midterm Solutions October 23, 2003 1. (a) Integrate by parts with u = x , dv = e 2 x dx : i xe 2 x dx = xe 2 x 2 - i e 2 x 2 dx = xe 2 x 2 - e 2 x 4 + C. (b) Easiest is the substitution x 2 + 9 = u : i 4 0 x 9 + x 2 dx = i x =4 x =0 u 1 / 2 2 du = u 1 / 2 v v v 2 9 5 = 5 - 3 = 2 . (c) Easiest is the substitution u = e x , but u = e x or u = e x + 1 is more natural. Using u = e x , du = e x dx : i 1 e x + 1 dx = i 1 u + 1 du e x = i 1 u ( u + 1) du = i p 1 u - 1 u + 1 P du = ln p u u + 1 P + C = ln p e x e x + 1 P + C = - ln(1 + e x ) + C. (d) Since the function f ( t ) = sin( t 3 ) is odd (that is, f ( - t ) = - f ( t )), the integral is zero. (e) By the Fundamental Theorem of Calculus g ( x ) = F (2003) - F ( x 2 ) where F ( t ) is an antiderivative of f ( t ) = sin( t 3 ). Thus g ( x ) = - F ( x 2 )( x 2 ) = - f ( x 2 )(2 x ) = - 2 x sin( x 6 ) . Alternatively, g ( x ) = - i x
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