MAT1332W10A1S

MAT1332W10A1S - MAT1332B 1332C Winter 2010 Assignment#1...

This preview shows pages 1–2. 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: MAT1332B, 1332C, Winter 2010, Assignment #1 Solutions 1. We use the substitution u = 1 + 4 t . Then du dt = 4, so dt = du 4 . Thus Z 1 1 + 4 t dt = 1 4 Z 1 u du = 1 4 ln | u | + C = 1 4 ln | 1 + 4 t | + C. (Don’t forget to resubstitute.) 2. Z 2- 2 ( y 3 + 4 y 5 ) dy = 1 4 y 4 + 4 6 y 6 ¶ | 2- 2 = 16 4 + 4 × 64 6 ¶- 16 4 + 4 × 64 6 ¶ = 0 . 3. Z π/ 2- π/ 2 [ x 2- 20sin( x )] dx = 1 3 x 3 + 20cos( x ) ¶ | π/ 2- π/ 2 = " π 3 24 + 0 #- "- π 3 24 + 0 # = π 3 12 4. First approach: First find indefinite integral by substitution u = 2 π ( x- 2). Then du dx = 2 π so dx = du 2 π . Hence Z cos(2 π ( x- 2)) dx = Z cos( u ) du 2 π = 1 2 π sin( u ) + C = 1 2 π sin(2 π ( x- 2)) + C. Then evaluate Z 5 2 cos(2 π ( x- 2)) dx = 1 2 π sin(2 π ( x- 2)) fl fl fl fl 5 2 = 1 2 π (sin(6 π )- sin(0)) = 0 . Second approach: Transform the limits of integration first. When x = 2, u = 2 π (2- 2) = 0. When x = 5, u = 2 π (5- 2) = 6 π . Then the integral after substitution becomes Z 5 2 cos(2...
View Full Document

{[ snackBarMessage ]}

Page1 / 3

MAT1332W10A1S - MAT1332B 1332C Winter 2010 Assignment#1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online