A1W08-138Sol

# A1W08-138Sol - Math 138 Assignment 1 Solutions Fall 2008 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 138 Assignment 1 Solutions Fall 2008 1. Evaluate each of the following indefinite integrals: a) Z x 2 ( x 3 + 5) 9 dx Solution Let u = x 3 + 5, then du = 3 x 2 dx , and by substitution, we obtain: Z ( u ) 9 du 3 = Z 1 3 u 9 du = 1 30 u 10 + C = 1 30 ( x 3 + 5) 10 + C . b) Z e x sin( e x ) dx Solution Let u = e x , then du = e x dx , and by substitution, we obtain: Z e x sin( e x ) dx = Z sin u du =- cos u + C =- cos( e x ) + C . c) Z tan- 1 x 1 + x 2 dx Solution Let u = tan- 1 x , then du = 1 1+ x 2 dx , and, by substitution, we obtain: Z tan- 1 x 1 + x 2 dx = Z u du = 1 2 u 2 + C = 1 2 (tan- 1 x ) 2 + C . d) Z x 1 + x 4 dx Solution Let u = x 2 , then du = 2 xdx , and, by substitution, we obtain: Z x 1 + x 4 dx = Z 1 1 + u 2 du 2 = 1 2 tan- 1 u + C = 1 2 tan- 1 ( x 2 ) + C . Math 138 Assignment 1 Solutions Fall 2008 2. Evaluate Z 1 2 sin- 1 x √ 1- x 2 dx . Solution 1 Let u = sin- 1 x , then du = 1 √ 1- x 2 , dx . Also u = 0 when x = 0, and u = π 6 when x = 1 2 . Sub- stitution leads to:...
View Full Document

## This note was uploaded on 04/07/2010 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

### Page1 / 6

A1W08-138Sol - Math 138 Assignment 1 Solutions Fall 2008 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online