MATH 206โ03 Quiz #4 Solutions
1. (8 points) Find the center of mass of the region bounded by the curves y=x3+ 2 and
y= 4x+ 2. Expressions need not be arithmetically reduced.
1
2
2
4
6
8
10
12
To find the center of mass, we need first find the area A,x-moment Mx, and y-moment
My:
A=Z2
0
(4x+ 2) โ(x3+ 2)dx =Z2
0
4xโx3= 2x2โx4
4๎2
0
= (8 โ4) โ(0 โ0) = 4
Mx=Z2
0
x[(4x+ 2) โ(x3+ 2)]dx =Z2
0
4x2โx4=4x3
3โx5
5๎2
0
=32
3โ32
5=64
15
My=Z2
0
1
2[(4x+ 2)2โ(x3+ 2)2]dx =Z2
0
โ1
2x6โ2x3+ 8x2+ 8xdx
=โx7
14 โx4
2+8x3
3+ 4x2๎2
0
=โ64
7โ8 + 64
3+ 16 = 424
21
So the center of mass is ๎Mx
A,My
A๎=๎16
15 ,108
21 ๎.
2. (8 points) Let f(x) = ๏ฃฑ
๏ฃฒ
๏ฃณ
0for x < 1
2
x3for xโฅ1
(a) (4 points) Verify that f(x)is a probability distribution function.
A cursory inspection reveals that this function is non-negative throughout: 0 is
non-negative everywhere, and 2
x3is non-negative as long as x > 0. The critical
property to demonstrate that this function is a probability distribution function
is simply that Rโ
โโ f(x)dx = 1. We can simplify this somewhat by ignoring the
region on which f(x) is zero, so that Rโ
โโ f(x) = Rโ
1f(x)dx. We evaluate this as
such:
Page 1 of 3 Thursday, October 16, 2008
