Math 231E. Homework 8. Solutions.
Problem 1. Show that the integral
Z
10
1
x2
dx ≤
4
2
x +x +1
10
5
by comparing it to a simpler integral.
Solution: First notice that x4 + x2 + 1 ≥ x4 for all x, and thus
x2
x2
1
≤
= 2.
4
2
4
x +x +1
x
x
So then
Z
5
10
x2
dx ≤
x4 + x2 + 1
10
Z
5
x=10
1
dx
1 1
1
= −
=− = .
2
x
x x=5
5 10
10
Problem 2. In each of the following problems, draw the region described and compute
its area.
a. The area below the curve y = x2 , above the x-axis, and between the lines x = 3 and
x = 6.
b. The area between the curves y1 = sin(x) and y2 = cos(x) and between the lines x = 0
and x = 2π.
c. The area below the x-axis and above the curve y = x2 − 2x − 1.
Solution:
R6
a. The area described is given by 3 x2 dx = 13 (216 − 27) = 63.
R 2π
b. The signed area is 0 sin(x) − cos(x)dx = 0, but the area between the curves is
Z
2π
|sin(x) − cos(x)| dx.
0
The total area between the curves can be found by choosing an interval where (say) sin(x) ≥
cos(x). From a graph such an interval is [ π4 , 5π
]. The area between the curves in this region
4
√
R 5π4
is π sin x − cos xdx = 2 2. There is a similar region where cos x ≥ sin x which is just
4
√
the reflection of the first region across x = 5π
. So the total area between the curves is 4 2.
4
c. This region
is bounded by the two roots of x2 − 2x − 1 = 0 which are x = 1 ±
√
√
R 1+ 2
is 1−√2 1 + 2x − x2 dx = 8 3 2 .
√
2. The area
Problem 3. Compute each of the following indefinite integrals:
Z
a.
sin(x) dx
Z
b.
x sin(x) dx
Z
c.
(x − 1)4
dx
x2
Solution:
Z
a. Standard:
sin(x) dx = − cos x + C
b. Integrate by Parts. Choose u = x, dv = sin(x) dx. This gives du = dx and v = − cos(x)
and we get
Z
Z
x sin(x) dx = −x cos(x) − (− cos x) dx = −x cos x + sin x + C.
c. Expand the numerator to find
Z Z 4
1
x − 4x3 + 6x2 − 4x + 1
4
2
dx =
x − 4x + 6 − + 2 dx
x2
x x
3
x
1
=
− 2x2 + 6x − 4 ln |x| − + C.
3
x
Problem 4. Compute each of the following indefinite integrals:
Z
Z
Z
sin(x)
a.
e
cos(x) dx
c.
x5 + 5x dx
e.
1
√
1 − x2 dx
1
√
x 1 − x2 dx
0
Z
b.
2 x
x e dx
Z
d.
Z
sin(x) dx
0
Solution:
π
f.
0
Z
a. u = sin(x) substitution gives
esin(x) cos(x) dx = esin x + C.
Z
b. Two integrations by parts shows
Z
c. Elementary
Z
d.
0
π
x2 ex dx = (x2 − 2x + 2)ex + C.
x5 + 5x dx = x6 /6 + 5x / ln 5 + C.
π
sin(x) dx = − cos x = 2.
0
Z
e. Trig sub x = sin t gives
1
√
1 − x2 dx =
0
f. u-sub u = x2 gives
Z
0
1
π
or realize as area of quarter-circle.
4
√
1
x 1 − x2 dx = .
3
Problem 5. In each of the following problems, you are given f (x), and you are supposed
to compute f 0 (x).
Z x
Z 1
2
−t2
a. f (x) =
e dt
e−t dt
c. f (x) =
0
Z
x
x2
b. f (x) =
e
−t2
Z
x2 sin(x)
2
e−t dt
d. f (x) =
dt
x cos(x)
0
Solution: These are all the fundamental theorem
Z x
2
2
e−t dt, so f 0 (x) = e−x
a. f (x) =
0
Z
x2
2
e−t dt, so f 0 (x) = 2xe−x
b. f (x) =
4
0
Z
c. f (x) =
1
2
e−t dt, so f 0 (x) = −e−x
2
x
Z
x2 sin(x)
d. f (x) =
x cos(x)
2
e−t dt gives f 0 (x) = (x2 cos x+2x sin x)e−x
4
sin2 x
−(cos x−x sin x)e−x
2
cos2 x
Problem 6. A farmer wants to build a fence around a field with maximal area, with the
following constraints: the field can either be in the shape of a rectangle or an isosceles
triangle, and one side of the field can be a river (and thus does not need to be fenced.) Let
us say that the total amount of material available for the fence is L.
a. Compute the maximal area obtainable if we assume that the farmer builds a rectangular field with one side along the river.
b. Compute the maximal area obtainable if we assume that the farmer builds a field
in the shape of an isosceles triangle, where the two equal sides are the fenced sides,
and the third side is the river.
c. Which of these two shapes is better?
Solution: They are (somewhat surprisingly) the same.
In the first (rectangular) case the field is L − 2x by x. The area is A = x(L − 2x). From there
dA
= L − 4x. This vanishes when x = L/4 giving the area as A = L2 /8. For completeness we
dx
should check the boundaries x = 0 and x = L but the area is obviously zero in those cases.
In the second (triangular) case
is x. Then
p the legs have length
√ L/2. Let us say that the base
√
1
1
2
2
2
2
the height is (by Pythagoras) L /4 − x /4 = 2 L − x . The area is A = 4 x L2 − x2 .
Computing
L2 − 2x2
dA
= √
.
dx
4 L 2 − x2
√
This vanishes when x = L/ 2, when the triangle is a right triangle. The area is again L2 /8.
Notice that, in each case, the field is a half of a square, bisected by the river.
Problem 7. Compute the limit
lim
x→5
1
x−5
Z
x
e dt .
t
5
Solution: By l’Hôpital and the fundamental theorem this is
Z x
1
ex
t
lim
e dt = lim
= e5 .
x→5
x→5 1
x−5 5