20b_03f_1s

20b_03f_1s - Math 20B First Midterm Solutions October 23,...

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

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/17/2009 for the course MATH 20B taught by Professor Justin during the Fall '08 term at UCSD.

Ask a homework question - tutors are online