Problem 1. a) Compute the definite integral
Z 2
4x
dx
2
0 x +2
This can be done by a u-substitution. Take u = x2 + 2, so that du = 2x dx, which menas
that 4x dx = 2 du. Notice that u(0) = 2 and u(2) = 6, so our integral becomes
Z 6
6
2
du = 2 ln |u| = 2 ln(6) − 2 ln(2) = 2 ln(6/2) = 2 ln(3).
2
2 u
b) Compute the integral
Z
x2 ln(x2 ) dx.
This one looks ripe for integration by parts. According to LAITE, we should take u = ln(x2 ).
3
Then du = 2x
dx = x2 dx. We need to use dv = x2 dx, so v = x3 . Then
x2
Z 3
Z
Z
x3
x3 ln(x2 )
x3 ln(x2 ) 2x3
2
2 2
x
2
2
2
dx =
−
x dx =
−
+ C.
x ln(x ) dx = ln(x )
−
3
3 x
3
3
3
9
c) Compute the integral
Z
π/6
0
sin 3x
dx
1 + cos 3x
First, let’s do the indefinite integral. This is a substitution. Use u = 1 + cos 3x, so that
du = −3 sin 3x dx. Then sin 3x dx = − 13 du. Rewriting the integral in terms of u, we get
Z
1
1
1 1
− · du = − ln |u| = − ln |1 + cos 3x| + C.
3 u
3
3
To evaluate the definite integral, we plug in the bounds
Z π/6
π/6
1
1
(1 − cos 3t) sin 3t dt = − ln |1 + cos 3x| = − (ln |1 + 0| − ln |1 + 1|)
3
3
0
0
1
ln 2
= − (ln 1 − ln 2) =
.
3
3
d) Compute
Z
x (ln x)2 dx
Hint: One strategy is to make a u-substitution and then integrate by parts.
Let’s try z = ln x, so that dz = x1 dx (I’m calling the new variable z because I want to reserve
u for the integration by parts).
dz = x1 dx and so x2 dz = x dx. But x2 = (ez )2 = e2z ,
R 2zThen
2
and so the integral becomes e z dz.
1
Now we need to do this integral, which is going to be by parts, twice. Take u = z 2 and
dv = e2z dz, so du = 2z dz and v = 12 e2z dz. Then we get
Z
Z
Z
1 2 2z
1
1 2 2z
2z 2
2z
e z dz = z e −
2ze dz = z e − ze2z dz
2
2
2
Do it again, now with u = z and dv = e2z dz, so du = dz and v = 12 e2z . Then we get
Z
Z
1 2z
1 2 2z
1 2 2z
1 2z
2z
z e − ze dz = z e −
ze −
e dz
2
2
2
2
1 2 2z
1 2z 1 2z
+C
= z e −
ze − e
2
2
4
1 2 1
1 2z
=
z − z+
e + C.
2
2
4
Now plug back in z = ln x and we get
Z
1
1
1
2
2
x (ln x) dx =
(ln x) − (ln x) +
x2 + C
2
2
4
(This is a tough one.)
2
Problem 2. a) Sketch and compute the area of the region bounded by y = 1 − x4 and y =
|x| − 1.
Here’s the region:
1.5
1.0
0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5
-0.5
-1.0
-1.5
The area is going to be
Z 1
Z
4
1
(1 − x ) − (|x| − 1) dx =
(2 − x4 − |x|) dx
−1
−1
The integrand is an even function (since it’s the sum of a bunch of even functions), and so
we can write this as
1
Z 1
x5 x2 1 1
13
4
−
2
(2 − x − x) dx = 2 2x −
=2 2− −
− 2(0 − 0 − 0) = .
5
2 0
5 2
5
0
b) Set up the integral for the arc length of the “top edge” of the region. (You don’t need to
actually evaluate it.)
The upper edge is defined by y = 1 − x4 , and the arc length is
Z 1p
Z 1p
Z 1√
0
2
3
2
1 + f (x) dx =
1 + (−4x ) dx =
1 + 16x6 dx.
−1
−1
−1
3
x
Z
cos t
dt.
t
Problem 3. Consider the function g(x) =
0
a) Is g(x) an increasing or decreasing function in the range 0 ≤ x ≤ π2 ? What about
π
≤ x ≤ π?
2
cos x
If x is in this range, then g 0 (x) =
is positive, and so the integral is increasing; thus
x
cos x
g(x) is an increasing function. On the second range, g 0 (x) =
is negative, which means
x
that g(x) is decreasing.
b) Compute the derivative
d
h (x) =
dx
0
Z
x
0
sin t
dt.
t
This is a matter for the fundamental theorem of calculus, which tells us that the answer is
sin x
simply
.
x
c) Compute the derivative
d
dx
Z
0
x2
sin t
dt.
t
This is equal to
d
−
dx
Z
0
x2
d
sin t
dt = − g(x2 ) = −g 0 (x2 )(2x),
t
dx
by the chain rule. We already found that g 0 (x) =
−
sin x
, and so the result is
x
2 sin x2
2x sin x2
=
−
x2
x
4
Problem 4. A parabolic antenna is formed by rotating the part of the graph of y = x2
between x = 0 and x = 3 around the y-axis.
a) Draw a sketch of the antenna, and compute its volume.
To find the volume, we slice the paraboloid horizontally. The area of the slice at height y is
√
given by A(y) = π y 2 = πy for 0 ≤ y ≤ 9, and so the volume is
Z
9
πy dy =
0
81π
.
2
b) A rainstorm fills the antenna with water to a height of y = 5 m. Assuming a density of
ρ = 1000 kg/m3 and gravitational force ρ = 9.8 m/s2 , set up the integral for the work done in
pumping the water out over the top of the antenna.
The water at height y needs to be lifted a distance of D(y) = 9 − y to get it over the rim of
the antenna. The work is then (using the notation from the book)
Z
b
Z
5
1000 · 9.8 · πy(9 − y) dy.
ρgA(y)D(y) dy =
a
0
The units of the answer are N · m.
5
2x
.
x2 − 3x − 10
The first order of business is to factor it: x2 −3x−10 = (x−5)(x+2). There are no repeated
or non-reducible factors, so we know that the final answer should take the form:
Problem 5. a) Compute the partial fraction expansion of
A
B
2x
=
+
x2 − 3x − 10
x−5 x+2
Clearing denominators yields
2x = A(x + 2) + B(x − 5)
Plugging in x = −2, we get −4 = B(−2−5), so B = 74 . Plugging in x = 5, we get 10 = A(7),
so A = 10
. This gives
7
2x
4
10
+
=
2
x − 3x − 10
7(x − 5) 7(x + 2)
b) Evaluate
Z
2x
dx.
x2 − 3x − 10
Using our answer to the previous problem, we obtain
4
10
ln |x − 5| + ln |x + 2| + C.
7
7
6
Problem 6. Sketch and find the area bounded by the curves y = −x, y =
√
x, and y = 2.
This is easier if we integrate with respect to y. The area is
2 3
Z 2
y 2 14
y
8 4
2
+
+
− (0 + 0) = .
(y − (−y)) dy =
=
3
2 0
3 2
3
0
Problem 7. a) Determine whether the improper integral
Z ∞
4dx
.
ex
1
is convergent or divergent.
We have
Z K
Z ∞
K
4 dx
−x
−x = lim
4e dx = lim −4e = lim −4e−K + 4e−1 = 4e−1 .
K→∞ 1
K→∞
K→∞
ex
1
1
The limit exists, so it’s convergent.
b) Determine whether the improper integral
Z 2
4x + 2
dx
2
1 x +x−2
is convergent or divergent.
To compute the integral, we need to have a partial fractions expansion. The denominator
factors as (x − 1)(x + 2), so
4x + 2
A
B
=
+
.
+x−2
x−1 x+2
x2
Clearing denominators, 4x + 2 = A(x + 2) + B(x − 1). Plugging in x = −2 we get −6 =
A(0) + B(−3), so B = 2. Plugging in x = 1 we get 6 = A(3), so A = 2. This means
4x + 2
2
2
=
+
.
+x−2
x−1 x+2
x2
Now for the actual integral. The integrand is not defined at x = 1, so we need a limit
Z 2
Z 2
Z 2
4x + 2
4x + 2
2
2
dx = lim
dx = lim
+
dx
2
2
K→1 K x + x − 2
K→1 K x − 1
x+2
1 x +x−2
2
= lim (2 ln |x − 1| + 2 ln |x + 2|) K→1
K
= lim (2 ln 1 + 2 ln(K + 2)) − (2 ln(K − 1) + 2 ln(K + 2))
K→1
The problem here is the term 2 ln(K − 1), which is going to be close to ln 0 when K is near
0: this causes the integral to be infinite.
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Problem 8. Let f (x) = x3 . Use the trapezoidal rule with n = 3 to estimate
Z 3
x3 dx.
0
We use the points 0, 1, 2, 3, so ∆x = 1. The formula says
Z 3
∆x
0 + 2 + 16 + 27
x3 dx ≈ (f (0) + 2 f (1) + 2 f (2) + f (3))
=
= 22.5
2
2
0
(The true value is 81/4 = 20.25, so this is a good approximation.)
Problem 9. Suppose you’re asked to estimate the volume of a football. You measure and
find that a football is 28 cm long. You use a piece of string and measure the circumference at
its widest point to be 53 cm. The circumference 7 cm from the end is 45 cm. Use Simpson’s
Rule to make your estimate.
Notice that 7 cm is a quarter the length of the football field, so we can use Simpson’s rule
with n = 4. The circumferences at different points are:
x 0 7 14 21 28
circ 0 45 53 45 0
The volume is
Z
45
A(x) dx
0
If the circumference is C, the radius is C/(2π), and the area A(x) is C 2 /(4π), and so the
areas are
x 0
7
14
21 28
2
2
2
A(x) 0 45 /(4π) 53 /(4π) 45 /(4π) 0
Then Simpson’s rule says
Z 45
7
A(x) dx ≈ (A(0) + 4 A(7) + 2 A(14) + 4 A(21) + A(28))
3
0
7
76363
= (0 + 4 · 452 /(4π) + 2 · 532 /(4π) + 4 · 452 /(4π) + 0) =
≈ 4051.18.
3
6π
This is in cubic centimeters. It sounds like a lot, but that’s roughly the same volume as a
cube with side length 15 cm, so it’s at least the right order of magnitude.
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