Math 172 Practice Test 1 Solutions
Warning. This has not been proofread. Use at your own peril. Feel free to inform me of any mistakes
you find. (I suspect that if there any mistakes they should be easy to spot and not out-weigh the benefit
of using this as a resource for studying.)
1. Find the area enclosed between y D x 2 and y D 2x C 3. Graph.
Solution. The graph is as such:
with y D x 2 in red and y D 2xC3. So (pretending you can’t just look on the graph and just find
Rb
out) we need to find the x-values in which these curves intersect and compute a .2xC3 x 2 /dx
(where a; b be are the x-values just mentioned).
x 2 D 2x C 3
!
x2
so a D 1 and b D 3. Now
Z 3
2
.2x C 3 x /dx D x 2 C 3x
1
3D0
2x
!
ˇ
1 3 ˇˇ3
x
D .9 C 9
3 ˇ 1
2. Give the Partial Fractions Decomposition of
xC4
:
x.x 2/2
1
3/.x C 1/ D 0
.x
9/
1
1
3C
3
D
32
:
3
Solution. We need to figure out what A; B, and C are if
A
xC4
B
C
D C
:
C
2
x.x 2/
x
x 2 .x 2/2
(1)
2/2 gives
Multiplying both sides by x.x
x C 4 D A.x
2/2 C Bx.x
2/ C C x:
So expanding everything
x C 4 D A.x 2
4x C 4/ C B.x 2
2x/ C C x
and regrouping as coefficients,
x C 4 D .A C B/x 2 C .C
4A
2B/x C .4A/:
Now, equating the coefficients gives us this system of equations:
0 D A C B;
1 D C 4A
4 D 4A:
(4) implies A D 1, then (2) implies B D
the expression in (1) gives
(2)
(3)
(4)
2B;
1, and finally (3) implies C D 3. Plugging this into
xC4
1
D
2
x.x 2/
x
1
x
2
C
3
.x
2/2
and we are done.
3.
a) State the Fundamental Theorem of Calculus (parts 1 & 2)
Solution.
(i) If f .x/ D
Rx
a
g.t/dt , then
f 0 .x/ D g.t/
(where a can be any constant number)
(ii) If there exists a function F .x/ such that F 0 .x/ D f .x/ (in other words, f has an
antiderivative), then
Z b
f .x/dx D F .b/ F .a/
a
b) Compute f 0 .x/ if f .x/ D
Rx
0
cos.t 2 /dt.
Solution. From direct use of the Fundamental theorem of Calculus (first part),
f 0 .x/ D cos.x 2 /
c) Compute g 0 .x/ if g.x/ D
R 2 ln x
1
e t =t dt
2
Solution. This is pretty much the same as
R ubefore. If the “2 ln x” scares you, make a
substitution u D 2 ln x. That way g.u/ D 1 e t =t dt, u0 D 2=x, and
eu
f .u/ D u0 D
u
0
4. Compute
R =2
0
e 2 ln x
2 ln x
2
2
D
D f 0 .x/
x
x ln x
cos.3x/ cos.5x/dx.
Solution. For this we’re going to have use partial fractions (twice ugh). Set
u D cos.3x/
dv D cos.5x/
so that
du D
Now,
vD
3 sin.3x/
1
3
cos.3x/ cos.5x/dx D cos.3x/ sin.5x/ C
5
5
Z
1
sin.5x/:
5
Z
sin.3x/ sin.5x/dx:
(5)
Next set
b
u D sin.3x/
db
v D sin.5x/
so that
db
u D 3 cos.3x/
and
b
vD
1
3
sin.3x/ sin.5x/ D sin.3x/ cos.5x/ C
5
5
Combining (5) and (6) gives
Z
Z
cos.3x/ cos.5x/dx D
1
cos.5x/
5
Z
cos.3x/ cos.5x/dx:
(6)
1
cos.3x/ sin.5x/
5
Z
3
3 1
sin.3x/ cos.5x/ C
cos.3x/ cos.5x/ dx :
C
5 5
5
This is equivalent to
Z
cos.3x/ cos.5x/dx D
1
cos.3x/ sin.5x/
5
Z
3
9
C
sin.3x/
: cos.5x/ C
cos.3x/ cos.5x/dx
25
25
R
Finally, solving for cos.3x/ cos.5x/ reveals
Z
5
3
cos.3x/ cos.5x/dx D
cos.3x/ sin.5x/ C
sin.3x/ cos.5x/ C C
16
16
3
and lastly (by making the observations cos.3=2/ D 0, cos.5=2/ D 0, and sin.0/ D 0),
ˇ=2
Z =2
ˇ
5
3
cos.3x/ cos.5x/dx D
cos.3x/ sin.5x/ C
sin.3x/ cos.5x/ˇˇ
D 0:
16
16
0
0
(I might have a mistake somewhere but I know the answer is supposed to be 0 and the method I
used is correct so it ended up not making a difference. Also, that problem was kind of difficult so
maybe there was an easier way to do this: one thing I noticed is that if you make the substitution
u D x =4, the integral turns into
Z =4
cos.3u C 3=4/ cos.5u C 5=4/du
=4
so if you could show the integrand is odd, i.e., f . x/ D x you’d be done. But that’s not
exactly easy to do—or as I found out in class today, you’ll get sheets with trig identities so you
can use one of the more obscure ones to make this way easier).
p
5. Find the volume generated when the area enclosed between y D x and y D x is revolved (I
think it’s revolved, can’t read the handwriting)
a) around the line y D
1 (use slicing).
p
Solution. By solving x D x, we find that the intersection points are x D 0 and x D 1.
Then the problem amounts to solving the integral
Z 1
Z 1
p
p
2
2
. x C 1/
.x C 1/ dx D x C 2 x C 1 x 2 2x 1 dx
0
0
ˇ1
ˇ
1 3
1 2 4 3=2
2
D x C x Cx
x
x
x ˇˇ
2
3
3
0
1 4
1
C C1
1 1 D
D
2 3
3
2
b) around the y-axis (use cylindrical shells)
Solution. With the cylindrical shells method, what p
we’re essentially doing is summing up
all cylinders which have radius x and height y D x (through integration). Then doing
the same thing for y D x and taking the difference as our answer. So
ˇ
Z 1
p
2 5=2 1 3 ˇˇ1
2 1
2
x ˇ D 2
D
2x. x x/dx D 2 x
5
3 0
5 3
15
0
6. Compute
R3
dx
0 .x 2 C9/3=2
Solution. This problem is screaming trig substitution. Recall the identity tan2 x C 1 D sec2 x.
If we substitute x D 3 tan u to compensate for the “9”, we have dx D 3 sec2 udu. Temporarily
ignoring the limits of integration,
Z
Z
Z
3 sec2 udu
1
sec2 udu
dx
D
D
p
p
3
3
.x 2 C 9/3=2
9
.3 tan u/2 C 9
sec2 u
Z
Z
1
du
1
D
D
cos udu D sin u C c: (7)
9
sec u
9
4
for some constant c. Now tan u D 3=x, so through basic trigonometry (tangent is opposite over
x
adjacent and sine being opposite over hypotenuse; sound familiar?) we have sin.u/ D p9Cx
2
so
Z
dx
x
1
D
Cc
p
9
.x 2 C 9/3=2
9 9 C x2
and therefore
ˇ3 Z 3
ˇ
1
dx
1
x
3
1
1
ˇ D
D
D
D
p
p
p
p
ˇ
2
3=2
9
9
9 C x2 0
18
3 18
9 2
0 .x C 9/
7. Let Sn D
3
n
Pn
i D1
2C
3i 2
n
a) Compute S3
Solution. Plugging in n D 3 gives
S3 D
n
X
.2 C i /2 D .2 C 1/2 C .2 C 2/2 C .2 C 3/2 D 9 C 16 C 25 D 50
iD1
b) Sn is a Riemann sum. What are the function f , the interval Œa; b, the partition P and the
points xi ?
Solution. By inspection, the interval width is 3, a D 2, f D x 2 , and the xi are the
left-hand endpoints. The partition is a regular partition, so it is being split evenly. Lastly
Z 5
lim Sn D
x 2 dx
n!1
2
c) Compute limn!1 SN (not by solving the previous integral).
Solution. Through properties of summations,
" n
#
n n n
n
X
3i 2
3X
12i
9i 2
3
12 X
9 X 2
3X
Sn D
2C
D
4C
C 2 D
4
1C
iC 2
i :
n i D1
n
n i D1
n
n
n
n
n
i D1
i D1
i D1
Recall the properties
n
X
iD1
So
1 D n;
n
X
i D1
iD
n.n C 1/
;
2
n
X
i D1
i2 D
n.n C 1/.2n C 1/
:
6
9
3
6n.n C 1/ 3n.n C 1/.2n C 1/
63
Sn D
4n C
C
D 2C
C 39
2
n
n
2n
2n
2n
So it is clear from this that Sn ! 39 as n ! 1 (which is the same thing you get upon
evaluating the integral).
R
8. Use Integration by parts to find a reduction formula for In D .ln x/n dx
5
Solution. Set
u D .ln x/n
du D
so that
n.ln x/n
x
Z
dv D dx
1
vDx
n
n
.ln x/ dx D x.ln x/
Z
n
.ln x/n 1 dx:
and we have the reduction formula
In D x.ln x/n
9. Let I D
intervals.
R4
2
dx
x
nIn
1
and let Tn be the approximation of I using the Trapezoid Method with n
a) Compute T4 .
Solution. (I’ve never heard of this before) The text states that the Trapezoidal method
defined by
b
Z
f .x/dx Tn D
a
x f .x0 / C 2f .x2 / C C 2f .xn 1 / C f .xn /
2
where x D .b a/=n and xi D a C ix. So in the case where a D 2; b D 4; f .x/ D
1=x, and n D 4 we have
Z 4
2
dx
T4 D f .2/ C 2f .2 C 2/ C 2f .2 C 4/ C 2f .2 C 6/ C f .2 C 8/
x
2
2
D f .2/ C 2f .4/ C 2f .6/ C 2f .8/ C f .10/
1 1 1 1
1
D C C C C
2 2 3 4 10
101
D
D 1:683N
60
b) Without using the exact value of I can you tell if TN < I ? Explain.
Solution. TN > I . f 00 .x/ D 2=x 3 > 0 for x in between 2 and 4, so f is concave up on
Œ2; 4 and therefore TN > I .
c) Estimate the error E4 D jI
T4 j.
Solution. Again, from the text, we need to find the maximum value of jf 00 .x/j. Now
f 00 .x/ D 2=x 3 , so f 000 .x/ D 6=x 4 . This has no critical points, so jf 00 .x/j must attain its
maximum at either 2 or 4. jf 00 .2/j D 1=4 and jf 00 .4/j D 1=32. Therefore jf 00 .x/j 1=4
for all x in the interval .2; 4/ and hence (from the text again)
jI
T4 j D jET j .1=4/.2/3
1
K.b a/3
D
D :
2
2
12n
12.4/
96
6