Math 2260 Exam #2 Practice Problem Solutions
1. Evaluate
Z
Answer: Recall that tan(x) =
sin(x)
cos(x) ,
tan3 (x) dx.
so the above integral is equal to
sin3 (x)
dx.
cos3 (x)
Z
Now, use the trig identity sin2 (x) = 1 − cos2 (x) to re-write this as
Z
sin2 (x) · sin(x)
dx =
cos3 (x)
(1 − cos2 (x)) sin(x)
dx
cos3 (x)
Z sin(x)
sin(x)
=
−
dx
3
cos (x) cos(x)
Z
Z
sin(x)
dx.
= tan(x) sec2 (x) dx −
cos(x)
Z
For the first, let u = tan(x). Then du = sec2 (x) dx and so
Z
Z
u2
tan2 (x)
tan(x) sec2 (x) dx = u du =
+ C1 =
+ C1 .
2
2
For the second integral above, let u = cos(x). Then du = − sin(x) dx, and so
Z
Z
sin(x)
du
=−
= − ln |u| + C2 = − ln | cos(x)| + C2 .
cos(x)
u
Combining these, then, we see that the given integral is equal to
tan2 (x)
+ ln | cos(x)| + C,
2
where I let C = C1 + C2 .
2. Integrate
Z
Answer: Using the trig identity sin2 θ =
Z
4
sin (2x) dx =
Z
2
sin (2x) dx =
2
sin4 (2x) dx.
1−cos(2θ)
,
2
Z we know that
1 − cos(4x)
2
2
Z
dx =
1 − 2 cos(4x) + cos2 (4x)
dx.
4
The first two terms will be easy to integrate, but for the third I need to use the power-reduction formula
again, so the above integral is equal to
Z 1
1 + cos(8x)
1
sin(4x) x sin(8x)
1 − 2 cos(4x) +
dx =
x−
+ +
+C
4
2
4
2
2
16
3x sin(4x) sin(8x)
=
−
+
+ C.
8
8
64
1
3. Integrate
Z √
x2 − 25
dx.
x
Answer: Use the trig substitution x = 5 sec θ. Then dx = 5 sec θ tan θ dθ, and so the above integral is
equal to
Z √
Z p
25 sec2 θ − 25
5 sec θ tan θ dθ = 5 sec2 θ − 1 tan θ dθ
5 sec θ
Z p
= 5 tan2 θ tan θ dθ
Z
= 5 tan2 θ dθ.
Z
sin2 θ
dθ.
=5
cos2 θ
This is a little annoying, but we can help ourselves out by using the trig identity sin2 θ = 1 − cos2 θ to
re-write it as
Z Z
1
1 − cos2 θ
dθ = 5
− 1 dθ
5
cos2 θ
cos2 θ
Z
=5
sec2 θ − 1 dθ
= 5 [tan θ − θ] + C.
Now, since x = 5 sec θ, we know√that sec θ = x/5, meaning that θ = sec−1 (x/5) and, after using the
2
Pythagorean Theorem, tan θ = x 5−25 . Therefore, our integral is equal to
#
"√
p
x2 − 25
−1
− sec (x/5) + C = x2 − 25 − 5 sec−1 (x/5) + C.
5
5
4. Evaluate
Z
e
∞
1
dx.
x(ln x)3
Answer: By definition
Z
∞
1
dx = lim
b→+∞
x(ln x)3
e
Z
e
b
1
dx,
x(ln x)3
which we can evaluate using the substitution u = ln x. Then du =
u(b) = ln(b), so the above limit is equal to
Z
lim
b→+∞
1
ln(b)
1
x
dx, u(e) = ln(e) = 1, and
−2 ln(b)
du
u
= lim
b→+∞ −2 1
u3
ln(b)
1
= lim − 2
b→+∞
2u 1
1
1
= lim −
+
b→+∞
2(ln b)2
2(1)2
1
=
2
since the first term goes to zero as b → +∞.
2
5. Suppose a 20 foot chain which weighs 5 pounds per foot is coiled on the ground. One end of the chain
is attached to a small crane. How much work does it take to lift this end 20 feet off the ground, so
that the chain is fully extended in the air?
Answer: Since the chain weighs 5 pounds per foot, when the end of the crane is x feet in the air, it
is supporting a weight of 5x, which is to say, exerting a force of F (x) = 5x. Hence,
20
Z
Z
20
5x2
5x dx =
2
F (x) dx =
W =
0
0
20
=
0
5(400)
− 0 = 1000,
2
so the crane is doing 1000 foot-pounds of work in lifting the chain until it is fully extended.
(Not accidentally, the chain weighs 5 · 20 = 100 pounds, and points on the chain are being lifted an
average of 10 feet off the ground; notice that 100 · 10 = 1000, which is the amount of work that we
calculated was done. Can you figure out why these two numbers coincide?)
6. Evaluate
Z
∞
√
0
dx
.
x(x + 4)
Answer: By definition,
∞
Z b
dx
dx
√
√
= lim
.
b→+∞
x(x
+
4)
x(x
+ 4)
0
0
√
√
√
1
Let u = x. Then du = 2√
dx, u(0) = 0 = 0, and u(b) = b, so the above limit is equal to
x
Z
b
Z
lim 2
b→+∞
0
√
1
1
· √ dx = lim 2
b→+∞
x+4 2 x
Z
b
0
1
du.
u2 + 4
Now, let u = 2 tan θ, meaning that θ = arctan(u/2). Therefore, θ(0) = arctan(0/2) = 0, θ(b) =
arctan(b/2), and du = 2 sec2 θ dθ, so the above is equal to
Z
arctan(b/2)
lim 2
b→+∞
0
1
4 tan2 θ + 4
· 2 sec2 θ dθ = lim
b→+∞
Z
arctan(b/2)
0
sec2 θ
dθ
sec2 θ
arctan(b/2)
= lim [θ]0
b→+∞
= lim (arctan(b/2) − 0)
b→+∞
=
π
,
2
since tan θ → +∞ as b → π/2− .
7. Solve the differential equation
(x2 + 1)
dy
= y.
dx
Answer: We want to separate variables and integrate, so we have
Z
Z
dx
dy
=
.
y
x2 + 1
The left hand side is easy; for the right hand side, let x = tan θ. Then dx = sec2 θ dθ, so
Z
Z
Z
dx
sec2 θ dθ
sec2 θ dθ
=
=
= θ + C.
2
2
x +1
sec2 θ
tan θ + 1
3
Therefore, since θ = arctan(x), we see that the above equation between integrals is equivalent to
ln y = arctan(x) + C.
Exponentiating both sides, we have that
y = earctan(x)+C = eC earctan(x) = Aearctan(x) ,
where I let A = eC .
8. Solve the differential equation
dy
= xy + x,
dx
y(0) = 10.
Answer: Notice that xy + x = x(y + 1), so separating variables and integrating yields
Z
Z
dy
= x dx
y+1
x2
ln(y + 1) =
+ C.
2
We can solve for y by exponentiating both sides
y + 1 = ex
2
/2+C
= eC ex
2
/2
2
= Aex
/2
(where A = eC ), and then subtracting 1 from both sides:
y = Aex
2
/2
− 1.
To determine A, we use the initial condition y(0) = 10:
y(0) = Ae0
2
/2
−1
10 = A − 1
11 = A.
Therefore,
2
y = 11ex
/2
− 1.
9. In a second-order chemical reaction, the reactant A is used up in such a way that the amount of it
present decreases at a rate proportional to the square of the amount present. Suppose this reaction
begins with 50 grams of A present, and after 10 seconds there are only 25 grams left. How long after
the beginning of the reaction will there be only 10 grams left? Will all of the A disappear in a finite
time, or will there always be a little bit present?
Answer: Since the rate of change of A is proportional to A2 , we have
dA
= kA2 .
dt
Since the reaction begins with 50 grams of A, we have A(0) = 50. Likewise, since there are 25 grams
left after 10 seconds, A(10) = 25. Then the first question is to determine the time t0 when A(t0 ) = 10,
and the second is to determine whether A(t) ever gets to zero.
To solve the differential equation, we separate variables and integrate:
Z
Z
dA
= k dt
A2
1
− = kt + C.
A
4
Therefore,
A(t) =
−1
.
kt + C
Now, we use the fact that A(0) = 50 to determine C:
−1
k(0) + C
−1
50 =
,
C
A(0) =
1
and thus A(t) =
so C = − 50
A(10) = 25:
−1
1
kt− 50
. Now we can determine the constant k using the fact that
−1
1
k(10) − 50
−1
25 =
1 .
10k − 50
A(10) =
Therefore,
1
25 10k −
50
250k −
= −1
1
= −1.
2
Hence
250k = −
1
2
1
and so k = − 500
. Therefore, the amount of A is determined by the equation
A(t) =
−1
t
− 500
−
1
50
=
−1
500
=
.
1
t + 10
− 500
(t + 10)
Now, we want to determine t0 so that A(t0 ) = 10:
500
t0 + 10
500
10 =
t0 + 10
A(t0 ) =
Therefore,
500
= 50,
10
so t0 = 50 − 10 = 40, meaning that there will be 10 grams left after 40 seconds.
t0 + 10 =
Finally,
500
t + 10
is always positive, so there will always be a little bit of A present.
A(t) =
10. On a hot day, a thermometer was brought outdoors from an air-conditioned building. The temperature
inside the building was 21◦ C, and so this is what the thermometer read at the moment it was brought
outside. One minute later the thermometer read 27◦ C, and a minute after that it read 31◦ C. What
was the temperature outside? (Impress us and express the answer without using logarithms or the
5
number e, but feel free not to work out products of two digit numbers. For example, if you find you
need to multiply 57 and 36 somewhere, you can just leave this as 57 · 36.)
Answer: By Newton’s Law of Cooling, if H(t) denotes the temperature of the thermometer after t
minutes, then
H(t) = Hs + H0 ekt ,
where Hs is the ambient temperature and H0 and k are constants. For this problem the goal is to
determine Hs .
Since the thermometer starts out at 21◦ C, we know that H(0) = 21:
H(0) = Hs + H0 ek(0)
21 = Hs + H0 ,
so H0 = 21 − Hs and so we can write
H(t) = Hs + (21 − Hs )ekt .
Since the temperature after 1 minute is 27◦ C, we have H(1) = 27:
H(1) = Hs + (21 − Hs )ek(1)
27 = Hs + (21 − Hs )ek ,
and so
ek =
Hence k = ln
27−Hs
21−Hs
27 − Hs
.
21 − Hs
, meaning that
27−Hs
H(t) = Hs + (21 − Hs )eln( 21−Hs )·t
t
ln 27−Hs
= Hs + (21 − Hs )e ( 21−Hs )
t
27 − Hs
= Hs + (21 − Hs )
.
21 − Hs
Finally, H(2) = 31, so we have
H(2) = Hs + (21 − Hs )
27 − Hs
21 − Hs
2
(27 − Hs )2
21 − Hs
Hs (21 − Hs ) 272 − 54Hs + Hs2
31 =
+
21 − Hs
21 − Hs
272 − 33Hs
31 =
.
21 − Hs
31 = Hs +
Therefore,
31(21 − Hs ) = 272 − 33Hs
31 · 21 − 31Hs = 272 − 33Hs
so
2Hs = 272 − 31 · 21
6
and
Hs =
272 − 31 · 21
.
2
If you happen to do the arithmetic, the right hand side is equal to
outside, which is pretty darn hot.
11. Evaluate
6
Z
4
729−651
2
=
78
2
= 39, so it is 39◦ C
x3 − 6x − 4
dx.
x2 − x − 6
Answer: Since there’s no obvious easier way of doing it, my plan is to use the method of partial
fractions. Since the degree of the numerator is greater than the degree of the denominator, my first
step should be to divide the numerator by the denominator
x2 − x − 6
x3 + 1 2
x + 0x − 6x − 4
x3 − x2 − 6x
x2 + 0x − 4
x2 − x − 6
x+2
Therefore,
Z
4
6
x3 − 6x − 4
dx =
x2 − x − 6
Z
4
6
x+2
x+1+ 2
x −x−6
dx.
Now, as I try to do partial fractions, notice that factoring the denominator makes things get really
easy:
x+2
1
x+2
=
=
,
x2 − x − 6
(x − 3)(x + 2)
x−3
so long as x 6= −2. Since −2 is not in the interval [4, 6], we don’t have to worry about it, so the integral
becomes
2
6
Z 6
1
x
x+1+
dx =
+ x + ln |x − 3|
x−3
2
4
4
36
16
=
+ 6 + ln(3) −
+ 4 + ln(1)
2
2
= (18 + 6 + ln(3)) − (8 + 4 + 0)
= 12 + ln(3).
12. What is the volume of the solid obtained by rotating the region between the graph of y =
the x-axis for 0 ≤ x ≤ 1 around the y-axis?
Answer: Here’s a graph of the function:
7
1
x2 +4x+3
and
0.5
0.4
0.3
0.2
0.1
0
0.25
0.5
0.75
1
-0.1
Certainly the shell method looks like the right way to go here. The shell through x will have radius
x − 0 = x, so the volume of the solid will be
Z 1
Z 1
x
1
dx = 2π
dx.
2πx 2
2 + 4x + 3
x
+
4x
+
3
x
0
0
This looks like a good candidate for the method of partial fractions, so first let me factor the denominator:
x
x
=
.
2
x + 4x + 3
(x + 3)(x + 1)
Then I guess that this fraction is the sum of two fractions
A
B
x
=
+
.
(x + 3)(x + 1)
x+3 x+1
Clearing denominators yields
x = A(x + 1) + B(x + 3)
x = (A + B)x + (A + 3B)
Then equating coefficients gives the system of equations
1=A+B
0 = A + 3B
The second equation says that A = −3B, so substituting that into the first yields
1 = −3B + B = −2B,
8
so B = −1/2 and hence A = 3/2. Therefore,
Z 1
Z 1
3/2
1/2
x
dx = 2π
−
dx
2π
x+3 x+1
0
0 (x + 3)(x + 1)
1
3
1
= 2π
ln(x + 3) − ln(x + 1)
2
2
0
1
3
1
3
ln(4) − ln(2) −
ln(3) − ln(1)
= 2π
2
2
2
2
= π (3 ln(4) − ln(2) − 3 ln(3))
= π ln(43 ) − ln(2) − ln(33 )
= π (ln(64) − ln(2 · 27))
64
= π ln
2 · 27
32
= π ln
.
27
Any of the answers in the last five lines would be fine.
13. Solve the initial-value problem
dy
dx
= ey sin x, y(0) = 0.
Answer: Since the differential equation is separable, I separate variables and integrate
Z
Z
dy
= sin x dx
ey
Z
Z
e−y dy = sin x dx
−e−y = − cos x + C,
and so e−y = cos x − C. Taking the natural log of both sides yields
−y = ln(cos x − C),
so we have that
y = − ln(cos x − C).
Using the initial condition, we have
y(0) = − ln(cos(0) − C)
0 = − ln(1 − C).
Since ln x = 0 only when x = 0, this means that C = 0, and so we see that
y = − ln(cos x),
which could also be written as ln (cos x)−1 = ln cos1 x = ln(sec x).
14. Evaluate
Z
e
x4 ln x dx.
0
Answer: Notice, first of all, than ln x is undefined at x = 0, so this is an improper integral. By
definition,
Z e
Z e
x4 ln x dx = lim
x4 ln x dx.
a→0+
0
9
a
I will integrate by parts using
u = ln x
1
dx
x
du =
dv = x4 dx
x5
5
v=
Then the above limit is equal to
5
e Z e 4 e
5
x
x5
x
x
lim
ln x −
dx = lim
ln x −
5
5
25 a
a→0+
a→0+
a 5
a
5
5
e5
a
a5
e
ln(e) −
−
ln(a) −
= lim+
5
25
5
25
a→0
5
e
e5
a5
a5
= lim+
−
−
ln(a) +
5
25
5
25
a→0
5
4e
a5
a5
= lim+
−
ln(a) +
25
5
25
a→0
4e5
1
=
−
lim a5 ln(a) + 0
25
5 a→0+
4e5
1
=
−
lim a5 ln(a).
25
5 a→0+
since lima→0+ a5 25 = 0.
Now, to evaluate the limit lima→0+ a5 ln(a), re-write as lima→0+
ln(a)
1
a5
, which is in the form
−∞
∞ ,
so we
can apply L’Hôpital’s Rule to get
lim+
a→0
1
a
−5
a6
= lim+
a→0
Therefore, we conclude that
a5
1 a6
·
= lim+
= 0.
a −5 a→0 −5
e
Z
4e5
.
25
x4 ln x dx =
0
15. Evaluate
Z
∞
√
1
Answer: By definition,
Z
1
∞
√
1 + x2
dx.
x6
1 + x2
dx = lim
b→∞
x6
Z
1
b
√
1 + x2
dx.
x6
To evaluate the integral, make the substitution x = tan θ. Then dx = sec2 θ dθ. Also, when 1 = x =
tan θ, we have that θ = π/4, and when b = x = tan θ, then θ = arctan(b). Therefore, the above limit
is equal to
Z arctan(b)
Z arctan(b) √
sec3 θ
1 + tan2 θ
2
sec
θ
dθ
=
lim
dθ
lim
b→∞ π/4
b→∞ π/4
tan6 θ
tan6 θ
Z arctan(b)
1
cos6 θ
= lim
·
dθ
b→∞ π/4
cos3 θ sin6 θ
Z arctan(b)
cos3 θ
= lim
dθ
b→∞ π/4
sin6 θ
10
Using the identity cos2 θ = 1 − sin2 θ, this is equal to
Z
arctan(b)
lim
b→∞
π/4
(1 − sin2 θ) cos θ
dθ = lim
b→∞
sin6 θ
Z
arctan(b)
π/4
cos θ
cos θ
−
sin6 θ sin4 θ
In turn, letting u = sin θ, we have that du = cos θ dθ and that u(π/4) =
sin(arctan(b)) =
√ b
,
b2 +1
Z
b→∞
√ b
b2 +1
−1 1
+
5
3
dθ.
and u(arctan(b)) =
so the above is equal to
lim
since limb→∞
√1
2
√
b
b2 +1
√
1/ 2
1
1
− 4
u6
u
√ b
b2 +1
−1
1
+
5
3
b→∞ 5u
3u 1/√2
1
−1
1
=
+
5u5
3u3 1/√2
du = lim
= 1 (use L’Hôpital!). Hence, the above is equal to
−
−1
1
√
√
+
5
5(1/ 2)
3(1/ 2)3
16. Evaluate
Z
3
4
2
=
−
15
√
√ !
√
2 2
−4 2 2 2
2
+
+
.
=
5
3
15
15
x
dx.
x2 − 6x + 5
Answer: I intend to use the method of partial fractions, so first factor the denominator
x2
x
x
=
.
− 6x + 5
(x − 5)(x − 1)
Now, write this fraction as the sum of two fractions
x
A
B
=
+
.
(x − 5)(x − 1)
x−5 x−1
Clearing denominators yields
x = A(x − 1) + B(x − 5)
x = (A + B)x + (−A − 5B).
Equating coefficients yields the system of equations
1=A+B
0 = −A − 5B.
From the second, I know that A = −5B. Substituting this into the first tells me that
1 = −5B + B = −4B,
11
so B = −1/4 and A = −5B = 5/4. Therefore,
Z 4
Z 4
5/4
1/4
x
dx =
−
dx
x−5 x−1
3 (x − 5)(x − 1)
3
4
1
= f rac54 ln |x − 5| − ln |x − 1|
4
3
5
1
5
1
=
ln(1) − ln(3) −
ln(2) − ln(2)
4
4
4
4
1
= − ln(3) − ln(2)
4
= ln 3−1/4 − ln(2)
−1/4 3
= ln
2
1
√
= ln
.
243
Any of the last four lines would be a reasonable answer.
17. Evaluate
π/8
Z
tan(4t) dt.
0
Answer: Re-write this integral as
Z
π/8
sin(4t)
dt.
cos(4t)
0
Notice that cos(4π/8) = cos(π/2) = 0, so this function is undefined at x = π/8. Therefore, this is an
improper integral, which by definition is equal to
Z b
sin(4t)
lim
.
−
b→π/8
0 cos(4t)
Letting u = cos(4t), we have that du = −4 sin(4t), and so the above is equal to
Z
Z
−1 b −4 sin(4t)
−1 cos(4b) du
lim −
dt = lim −
4 0 cos(4t)
4 1
u
b→π/8
b→π/8
−1
cos(4b)
[ln |u|]1
= lim −
4
b→π/8
−1
= lim −
(ln(cos(4b)) − ln(1))
4
b→π/8
−1
= lim −
ln(cos(4b)).
4
b→π/8
But this limit is equal to +∞, so the integral diverges.
18. Evaluate
2
Z
1
ln x
dx.
x2
Answer: I will integrate by parts using
u = ln x
du =
1
dx
x
12
1
dx
x2
−1
v=
,
x
dv =
so the above integral is equal to
2 Z 2
2
− ln x
− ln 2 ln 1
1
−1
dx
=
+
+
+
2
x
2
1
x 1
1 x
1
−1
− ln 2
+
+1
=
2
2
1
= (1 − ln 2).
2
19. Evaluate
π/2
Z
x2 cos(x) dx.
0
Answer: I will integrate by parts using
u = x2
dv = cos(x) dx
du = 2x dx
v = sin(x),
so the above integral is equal to
2
π/2
x sin(x) 0 −
Z
0
π/2
Z π/2
π2
(1) − 0 − 2
x sin(x) dx
2x sin(x) dx =
4
0
Z π/2
π2
=
−2
x sin(x) dx.
4
0
To deal with the remaining integral, I integrate by parts again using
u=x
du = dx
dv = sin(x) dx
v = − cos(x),
so the above is equal to
π2
π/2
− 2 [−x cos(x)]0 +
4
!
π/2
Z
cos(x) dx
=
0
π2
π/2
− 2 [−x cos(x) + sin(x)]0
4
π2
− 2 (0 + 1 + 0 − 0)
4
2
π
− 2.
=
4
=
20. Evaluate
√
Z
π
x sin2 x2 dx.
0
√
Answer: Let u = x . Then du = 2x dx and, since u(0) = 0 and u( π) = π, the above integral is
equal to
Z
Z
1 π 2
1 π 1 − cos(2u)
sin (u) du =
du
2 0
2 0
2
π
1 u sin(2u)
=
−
2 2
4
0
h
i
1 π
− 0 − (0 − 0)
=
2
2
π
= .
4
2
13