# 18107 - on 0 ≤ t ≤ π is 1 π Z π sin tdt = 1 π [-cos...

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

7.1, #21. Z ± 3 t - 2 t 2 ² dt = 3ln | t | + 2 t - 1 + C 7.1, #26. Z ( e x + 5) dx = e x + 5 x + C 7.1, #43. The indeﬁnite integral is 7.1, #14 on assignment 6. Z 2 1 1 + y 2 y dy = (ln y + y 2 / 2) ³ ³ 2 1 = ln2 + 3 / 2 = 2 . 193 7.1, #54. The average value of sin t on the interval 0 t 2 π is 1 2 π - 0 Z 2 π 0 sin tdt = 0 . We can evaluate this integral using an antiderivative or by knowing that the areas above and below the x -axis are equal. This is becasue sin( t + π ) = - sin t . The part of the graph above the x -axis balances the part below. This is a huristic argument that the average should be 0. The average value of sin t
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: on 0 ≤ t ≤ π is 1 π Z π sin tdt = 1 π [-cos t ] π = 1 π (-(-1)-(-1)) = 2 π . 7.2, #2. Use the substitution u = x 2 , du = 2 xdx to get Z 2 x cos( x 2 ) dx = Z cos udu = sin u + C = sin( x 2 ) + C 7.2, #5. Use the substitution u = sin x , du = cos xdx to get Z e sin x cos xdx = Z e u du = e u + C = e sin x + C 7.2, #6. Use the substitution u = x 2 + 1, du = 2 xdx to get Z x x 2 + 1 dx = 1 2 Z du u = 1 2 ln | u | + C = 1 2 ln( x 2 + 1) + C...
View Full Document

## This note was uploaded on 07/06/2011 for the course MATH 180 taught by Professor Tan during the Fall '08 term at Ill. Chicago.

Ask a homework question - tutors are online