Problem 1.
f (x) = 5 − x2 ,
g(x) = 3 − x.
a) Find the x-coordinate of the intersection points of the two functions.
To find the intersections, we need to solve 5 − x2 = 3 − x. This is x2 − x − 2 = 0,
(x − 2)(x + 1) = 0, which means that x = −1 or x = 2.
b) Compute the area of the region bounded by the graphs of f (x) and g(x).
It always helps to draw the region, so we can see what are the top and bottom functions.
4
2
-2
-1
1
2
3
-2
-4
The area is given by
Z 2
Z 2
(upper − lower) dx =
(5 − x2 ) − (3 − x) dx
−1
−1
Z 2
x2 x3 2
2
−
=
2 + x − x dx = 2x +
2
3
−1
−1
9
= (4 + 2 − 8/3) − (−2 + 1/2 + 1/3) = .
2
Problem 2. Evaluate the following integrals. Be sure to specify which integration methods
you are using.
Z
√
a) x 1 − x dx.
There are a couple ways to do this one. The easiest is by substitution: take u = 1 − x, so
du = −dx. Then x = 1 − u, and the integral becomes
Z
Z
Z
√
√
√
2
2
x 1 − x dx = − (1 − u) u, du = −
u − u3/2 du = − u3/2 + u5/2 + C
3
5
2
2
= − (1 − x)3/2 + (1 − x)5/2 + C
3
5
√
Let’s try integration by parts instead. Take u = x and dv = 1 − x dx. This gives du = dx
1
and v = − 23 (1 − x)3/2 . Then
Z
Z
√
2
2
3/2
x 1 − x dx = − x(1 − x) − − (1 − x)3/2 dx
3
3
Z
2
2
= − x(1 − x)3/2 +
(1 − x)3/2 dx
3
3
2
2
2
3/2
5/2
+C
− (1 − x)
= − x(1 − x) +
3
3
5
2
4
= − x(1 − x)3/2 − (1 − x)5/2 + C.
3
15
With a bit of algebra, you can check that this is the same answer we get with substitution.
b)
Z
x2 e−2x dx.
We’ve dealt with this one before: it’s a problem where you need to use integration by parts
twice. The first time, take u = x2 and dv = e−2x dx. Then du = 2x dx and v = − 12 e−2x .
Integration by parts gives
Z
Z
1
1 2 −2x
2 −2x
− − e−2x 2x dx
xe
dx = − x e
2
2
Z
1
= − x2 e−2x + xe−2x dx.
2
We need to use integration by parts again on the remaining integral. Take u = x and
dv = e−2x dx. Then du = dx and v = − 12 e−2x dx. We get
Z
Z
1 2 −2x
2 −2x
+ xe−2x dx
xe
dx = − x e
2
Z
1 2 −2x
1 −2x
1 −2x
=− x e
+ − xe
− − e
dx
2
2
2
Z
1 2 −2x 1 −2x 1
=− x e
− xe
+
e−2x dx
2
2
2
1 2 −2x 1 −2x 1 −2x
=− x e
− xe
− e
2
2
4
Problem 3. a) Write the general formula for the length of the graph of a function f (x)
between x = a and x = b.
The formula is
Z bp
L=
1 + f 0 (x)2 dx.
a
2
b) Find the length of the graph of the function f (x) = 2 ln(x) − x16 between x = 1 and x = 6.
2
We have f 0 (x) =
2
x
− x8 , so
4
1 x2
+
−
x2 2 64
4
1 x2
1 + f 0 (x)2 = 2 + +
x
2 64
2
2 x
+
=
x 8
p
2 x
1 + f 0 (x)2 = + .
x 8
f 0 (x)2 =
Then the length is
Z 6
2 x
x2 6
35
+ dx = 2 ln |x| + = (2 ln 6 + 36/16) − 0 + 1/16) = 2 ln 6 − .
8
16 1
16
1 x
Problem 4. Let R be the region in the first quadrant bounded by the graphs of the functions
f (x) = 2x2 − 1 and g(x) = x, and consider the solid of revolution obtained by revolving R
around the y-axis.
a) Compute the volume V of this solid of revolution using slices (these may be disks or
washers).
The region is question is shaded.
1.0
0.5
0.2
0.4
0.6
0.8
1.0
-0.5
-1.0
We’re rotating about the y-axis, so we want toqdo an integral dy. The larger radius is the
curve y = 2x2 − 1, which gives a radius of x = y+1
. The inner radius is x = y.
2
Thus we need to compute
 r

!2
y+1
− (y)2  dy
V =
π
2
0
Z 1
y+1
2
=π
−y
dy
2
0
2
1
y
y y 3 + −
=π
4
2
3 0
5π
1 1 1
=π
+ −
− (0 + 0 + 0) =
.
4 2 3
12
Z
1
3
b) Compute the volume V of this solid of revolution using shells.
We didn’t cover this, so don’t worry about it.
Problem 5. Evaluate the following integrals. Specify the method you used.
Z
a) (ln(x))2 dx.
There’s no obvious useful substitution, so this looks like fodder for integration by parts.
Take u = (ln x)2 and dv = dx, so du = 2 ln x x1 dx and v = x. Then
Z
Z
1
2
2
(ln(x)) dx = x(ln x) − x2 ln x dx
x
Z
= x(ln x)2 − 2 ln x dx
This is progress – we know how to do this integral. Integrate by parts a second time, now
with u = ln x and du = x1 dx and dv = dx, so v = x. Then
Z
Z
2
2
(ln(x)) dx = x(ln x) − 2 ln x dx
Z
1
2
= x(ln x) − 2 x ln x − x dx
x
Z
2
= x(ln x) − 2 x ln x − 1 dx
= x(ln x)2 − 2 (x ln x − x) + C
= x(ln x)2 − 2x ln x + 2x + C.
Z
b)
√
cos( x) dx.
√
Let’s first try a substitution z = x; I’m calling it z just in case we need to later integrate
by parts, and want to call something else u – you never know. Then dz = 2√1 x dx, which
√
means dx = 2 x dz. The integral becomes
Z
Z
Z
√
cos( x) dx = 2z cos(z) dz = 2 z cos(z) dz.
Now integrate by parts. Take u = z and dv = cos z dz, so du = dz and v = sin z. Then
Z
Z
Z
√
cos( x) dx = 2 z cos(z) dz = 2 z sin z − sin z dz = 2 (z sin z + cos z) .
Plugging back in z =
√
x and remembering the constant, our answer is
Z
√
√
√
√
cos( x) dx = 2 x sin x + 2 cos x + C.
4
Problem 6. Consider the region bounded by y = x2 , the y-axis, and y = 4. Find the volume
of revolution of the resulting solid, when it is rotated about:
a) The y-axis.
√
We’re rotating about the y-axis, so we need an integral dy. The radius is x = y, and so
the volume is
2 4
Z 4
Z 4
y √ 2
y dy = π
= 8π.
π( y) dy = π
2 0
0
0
b) The axis y = 5.
This time, we can use the washer method by doing an integral dx. The upper bound on x
is x = 2 (corresponding to y = 4). The bigger radius is the distance from from y = 5 to
y = x2 , which is 5 − x2 . The smaller radius is the distance from y = 5 to y = 4, which is 1.
So the volume is
2
Z 2
Z 2
10x3 x5 416π
2
4
2 2
+
.
(24 − 10x + x ) dx = π 24x −
=
π (5 − x ) − 1 = π
3
5 0
15
0
0
5