Math 106 Solutions
Quiz 2
9/21/12
1. Let I=Zπ/2
0
xsin x dx. How many subdivisions are required to obtain a left sum approximation
with error of at most 1/10,000?
Since f(x) = xsin x, then f0(x) = sin x+xcos x.
To find K1we need to find the maximum value
of |sin x+xcos x|on [0,π
2]. Looking at the
graph reveals that |f0(x)|achieves a maximum
value of approximately 1.391008 on [0,π /2], so
let K1= 1.4. The error bound estimates for left
sums may be determined using:
|I−Ln| ≤ K1(b−a)2
2n.
0.5
1.0
1.5
0.2
0.4
0.6
0.8
1.0
1.2
1.4
| f’ |
Therefore,
K1(b−a)2
2n≤1
10000 ⇐⇒ 1.4π
2−02
2n≤1
10000
⇐⇒ n≥14000π2
8≈17271.81.
Therefore, we require (at least) n= 17272 subdivisions.
2. To find the area of the region between y= 2x2
and y=x4−2x2, first find the x-coordinates for
the intersection points:
2x2=x4−2x2⇐⇒ x4−4x2= 0
⇐⇒ x2(x2−4) = 0
⇐⇒ x2(x+ 2)(x−2) = 0
⇐⇒ x=−2,0,2
-2 0 2
y= 2x2
y=x4−2x2
Notice that 2x2≥x4−2x2on [−2,2], therefore we may view y= 2x2as the “top” function and
y=x4−2x2as the “bottom” function. In addition, the shaded region is symmetric across the
y−axis, therefore we can find the area on the right hand side (i.e., from x= 0 to x= 2) and double
it to find the desired area. The area of the shaded region may be found using the integral below:
2Z2
0
[2x2−(x4−2x2)] dx = 2 Z2
0
(4x2−x4)dx = 2 4
3x3−1
5x52
0
= 2 4
3(2)3−1
5(2)5−0
= 2 32
3−32
5=128
15
1
