MTH 240 Final: MTH240 Term Test 1 2011 Winter
Ryerson University
Department of Mathematics
MTH 240 Winter 2011 – Test I
LAST NAME: FIRST NAME:
(Please print) (Please print)
I.D. NUMBER: SIGNATURE:
Date: February 18, 2011 Duration: 90 minutes
Professor (circle one)
Kim Poliakov
Tasi´c Wang
Section:
INSTRUCTIONS:
•Verify that your exam has SIX (6) pages in-
cluding this page.
•The use of notes, formula sheets, books or cal-
culators is not allowed.
•The tests written in pencil or erasable
pens are not eligible for remarking.
•For full-answer questions:
Give full justification for your answers; correct
answers alone may be worth nothing. Cross out
or erase all rough work not relevant to your so-
lution. Write your solutions in the space pro-
vided. If you need more space, use the back of
the page. Indicate this fact on the original page,
making sure that your solution cannot be con-
fused with any rough work which may be there.
For markers’ use only:
Page Value Mark
2 10
3 10
4 10
5 10
6 10
Total 50
1
MTH 240 Test I
1. (5+5 pts.) Evaluate the following indefinite integrals.
(a) Zsin(√x)dx
Soln: Make a substitution: y=√x. So, dy =1
2√xdx.
Zsin(√x)dx =Z2ysin y dy
Let u= 2y, dv = sin ydy. Then du = 2dy, v =−cos y
Z2ysin y dy = (2y)(−cos y)−Z(−cos y)(2dy)
=−2ycos y+ 2 sin y+C
So, Zsin(√x)dx =−2√xcos(√x) + 2 sin(√x) + C
(b) Z(sec x)4dx
Soln:
Z(sec x)4dx =Ztan2x+ 1sec2xdx
=Z(tan2x)(sec2x)dx +Zsec2xdx
=Z(tan2x)(sec2x)dx + tan x
For Z(tan2x)(sec2x)dx, we let u= tan x. So du = sec2xdx, and
Z(tan2x)(sec2x)dx =Zu2du
=1
3u3+C
=1
3tan3x+C
So, Z(sec x)4dx =1
3tan3x+ tan x+C
2
Document Summary
Give full justi cation for your answers; correct answers alone may be worth nothing. Cross out or erase all rough work not relevant to your so- lution. Write your solutions in the space pro- vided. If you need more space, use the back of the page. Indicate this fact on the original page, making sure that your solution cannot be con- fused with any rough work which may be there. Evaluate the following inde nite integrals: (5+5 pts. ) (a) sin, dx. Soln: make a substitution: y : so, dy = 1. 2 (cid:90) sin( xdx. (cid:90: dx = Let u = 2y, dv = sin ydy. Then du = 2dy, v = cos y (cid:90) (cid:90) ( cos y)(2dy) 2y sin y dy = (2y)( cos y) (cid:90) So du = sec2 xdx, and (tan2 x)(sec2 x)dx = (cid:90) u2du u3 + c tan3 x + c. 3 tan3 x + tan x + c.
