MATH 106A,B - CALCULUS II WINTER 2009
QUIZ 7
NAME:
Show ALL your work CAREFULLY.
(a) Evaluate the improper integral
Z1
0
1
3
p(3xโ1) dx.
Using a simple substitution with u= 3xโ1, the indefinite integral
Z1
3
p(3xโ1) dx =Zuโ1/31
3du
=1
3
u2/3
(2/3) +C=1
2(3xโ1)2/3+C.
Now, the improper integral can be written as
Z1
0
1
3
p(3xโ1) dx =Z1/3
0
1
3
p(3xโ1) dx +Z1
1/3
1
3
p(3xโ1) dx
= lim
aโ1/3Za
0
1
3
p(3xโ1) dx + lim
bโ1/3Z1
b
1
3
p(3xโ1) dx
= lim
aโ1/3๎1
2(3aโ1)2/3โ1
2(โ1)2/3๎+ lim
bโ1/3๎1
2(3 โ1)2/3โ1
2(3bโ1)2/3๎
=1
2(3
โ4โ1).
(b) Use comparison to determine whether the improper integral
Zโ
2
1
x+x4dx
converges or diverges. Justify your answer.
For xโฅ2,x +x4> x4so that 0<1
x+x4<1
x4. It follows that
0<Zโ
2
1
x+x4dx < Zโ
2
1
x4dx < Zโ
1
1
x4dx < โ.
The last inequality is the result of the p-test since p= 4 >1. Thus, by comparison, we have
shown that the improper integral Rโ
2
1
x+x4dx converges.
Date: March 11, 2009.
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