MATH 106A,B - CALCULUS II WINTER 2008
QUIZ 7
NAME:
Show ALL your work CAREFULLY.
(a) Let kbe a non-zero constant. Determine whether the improper integral
Z∞
0
kxe−x2dx
converges or diverges. [Hint: evaluate it!]
Let u=−x2so that du =−2xdx. Then,
Zxe−x2dx =−1
2Zeudu =−1
2e−x2+C.
Now, we have
Z∞
0
kxe−x2dx =klim
b→∞ Zb
0
kxe−x2dx
=klim
b→∞
−1
2e−x2
b
0
=klim
b→∞ −1
2e−b2+1
2=k
2<∞.
So the integral converges.
(b) Use comparison to determine whether the improper integral
Z∞
2
1
1 + x3dx
converges or diverges. Justify your answer.
Note that 1
1+x3<1
x3for x≥2so
0<Z∞
2
1
1 + x3dx < Z∞
1
1
1 + x3dx < Z∞
1
1
x3dx < ∞
since R∞
1
1
xpdx converges for p > 1. It follows that the improper integral converges.
(c) Let
f(x) = (kxe−x2,if x≥0;
0,otherwise.
Based upon your calculation in (a), find the value of k, if it exists, for which the function fa probability
density function.
For fto be a p.d.f., R∞
−∞ f(x)dx = 1 and f(x)≥0for all x. Note that R∞
−∞ f(x)dx =R∞
0f(x)dx =k
2,
by (a). Thus, when k= 2, the improper integral is equal to 1and f(x)≥0. Hence fis a p.d.f.
for k= 2.
Date: March 5, 2008.
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