Math 172-507 Practice Test 2 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. Write out the form of the partial fraction decomposition. Do not compute coefficients:
x2 C x C 1
x 2 .x 2 1/.x 2 C 1/
Solution. Note that x 2
1 D .x
1/.x C 1/. Then
x2 C x C 1
A
B
C
D
Ex C F
D C 2C
C
C 2
:
2
2
2
x .x
1/.x C 1/
x
x
x 1 xC1
x C1
(1)
(From here one would multiply both sides of the equation in (1) by the denominator and equate
coefficients in order to get a system of equations. Once A; B; C; D; E; and F are solved for,
simply plug them into (1) to get the final result.)
==
2.
a) Use the trapezoid formula with n D 4 to approximate
R4
dx
2 2x 1 .
Solution. Recall the Trapezoidal Rule:
Z
b
f .x/dx Tn D
a
x f .x0 / C 2f .x1 / C 2f .x2 / C C 2f .xn 1 / C f .xn / ; (2)
2
where x D .b a/=n and xi D a C ix. In this case, we have a D 2, b D 4,
f .x/ D 1=.2x 1/, and n D 4. Then x D 1=2 and so xi D 2 C i.1=2/. Using this
information in (2) yields
Z 4
1
dx
T4 D Œf .2/ C 2f .5=2/ C C2f .3/ C 2f .7=2/ C f .4/
1
4
2 2x
1
D Œ1=3 C 1=2 C 2=5 C 1=3 C 1=7
4
359
D
:
==
840
b) Estimate the error.
Solution. Recall the error bound for the Trapezoid Rule:
jET j 1
K.b a/3
;
12n2
(3)
where jf 00 .x/j K for a x b, and ET is the error in the Trapezoidal Rule. In this
case we have a D 2; b D 4, n D 4, and f 00 .x/ D 8=.2x 1/3 . 2 x 4, so
ˇ
ˇ
ˇ
ˇ 00 ˇ ˇ
8
8
ˇD
ˇf .x/ˇ D ˇ
(4)
ˇ .2x 1/3 ˇ .2x 1/3
(the last equality coming from the fact that .2x 1/3 > 0 when 2 x 4). Consider the
denominator in (4): .2x 1/3 . Because it’s in the denominator, the smaller it is, the larger
the value of (4) will be. Since we are looking for the largest value of (4) on 2 x 4,
we must find the smallest value of .2x 1/3 . .2x 1/3 is increasing, so the smallest
value of .2x 1/3 on 2 x 4 is the output of the smallest x on 2 x 4, i.e.,
.2.2/ 1/3 D 27. Therefore
ˇ 00 ˇ
ˇf .x/ˇ D
8
1/3
.2x
8
27
and K D 8=27. Putting all of these into (3) gives
K.b a/3
.8=27/.4 2/3
1
jET j D
D
12n2
12.4/2
81
==
c) Can you tell if your approximation is an overestimate or an underestimate of the integral?
Explain.
Solution. Let f .x/ D 1=.2x
1/. From the previous part,
f 00 .x/ D
8
.2x
1/3
>0
when 2 x 4, so f .x/ is concave up on this interval. Therefore the approximation
is an overestimate of the integral. (If not convinced, observe the following graph of the
Trapezoid Rule on an arbitrary concave up function:
From this, it is easy to see that the Trapezoidal Rule is an overestimate on concave up
functions).
==
2
3. Let C be the arc y D x 2 =2, 0 x 2.
a) Write a definite integral for the length of C . (Do not compute the integral.)
Solution. Recall the arc length formula for a function y D f .x/:
s
2
Z 2
dy
dx:
LD
1C
dx
0
(5)
In our case we have f .x/ D y D x 2 =2. Then
dy
Dx
dx
and so
dy
dx
2
D x2:
Using this information in (5) yields
2
Z
LD
p
1 C x 2 dx:
(6)
==
0
b) Compute the surface area generated when C is revolved about the y-axis.
Solution. Recall the formula for the surface area generated when a curve C is revolved
around the y-axis:
Z
S D 2x ds;
where ds is
s
dy
1C
dx
s
2
dx
or
dx
1C
dy
2
dy:
Note that the former is the expression in (6). So
Z 2
p
SD
2x 1 C x 2 dx:
0
Making the u-substitution u D x 2 , we have that du D 2xdx and so
Z 2
p
SD
2x 1 C x 2 dx
0
Z 4
p
1 C udu
D
0
ˇ4
2 h
3=2 ˇ
D
.u C 1/ ˇ
0
3
2 p
D
5 5 1
3
21:3:
==
4. Find the coordinates .x;
N y/
N of the centroid of the region enclosed between y D x 2 =2 and the
line y D 2.
3
Solution. Recall the formulas for xN and yN for region which lies between two curves y D f .x/
and y D g.x/:
Z
1 b xN D
x f .x/ g.x/ dx
(7)
A a
and
Z
2 2 o
1 b 1 n
f .x/
g.x/ dx;
(8)
yN D
A a 2
where A is the area of the region. To find A, we must find a and b of this region, i.e., where
y D 2 and y D x 2 =2 intersect. We compute
x2
D2
2
)
so that
Z
2
AD
2
2
x2 D 4
x2
dx D 2x
2
)
x D ˙2
ˇ2
x 3 ˇˇ
16
D :
ˇ
6 2
3
Using this information in (8) gives
2 Z 2 3
1
x
2
yN D
.2/
dx
16 2 2
2
Z 2 1
x4
3
4
dx
D
16 2 2
4
ˇ2
3
x 5 ˇˇ
6
D
4x
D :
ˇ
32
20 2
5
Observe that the region under consideration is symmetric about the y-axis, so xN D 0. Therefore
.x;
N y/
N D .0; 6=5/.
==
R 1 dx
is improper. Write the proper definition. Does the integral converge? If so
5. Explain why 0 p
3
x
compute its value. If not explain why.
Solution. We separate the problem into three parts.
p
(i) the integral is improper because limx!0C 1= 3 x D 1.
(ii) The definition of a proper integral is the following:
“An integral which has neither limit infinite and from which the integrand does
not approach infinity at any point in the range of integration.”
p
(iii) 1= 3 x is defined for 0 < x 1, so we compute
Z 1
Z 1
dx
3 h 2=3 ˇˇ1
3
3
1=3
x ˇ D
lim x 2=3 D
D
x
dx
D
p
3
0
2
2 x!0C
2
x
0
0
p
so the integral converges. (We use limx!0C x 2=3 as a technicality; 1= 3 x is not defined at
x D 0 so we use a limit to evaluate its integral—even though its antiderivative is defined
there.)
==
6.
a) State the comparison theorem for improper integrals of the form
4
R1
a
f .x/dx.
Solution. The comparison theorem for improper integrals of the form
stated as follows:
R1
0
f .x/dx is
Suppose that f and g are continuous functions with f .x/ g.x/ 0 for x a.
R1
R1
(i) If a f .x/dx is convergent, then a g.x/dx is convergent.
R1
R1
(ii) If a g.x/dx is divergent, then a f .x/dx is divergent.
==
b) Use a) to show that
R1
1
p dx
x 3 C2
converges.
p
p
Solution. The square root function is increasing, so x 3 x 3 C 2. Taking reciprocals
yields the inequality
1
1
p :
p
x3 C 2
x3
R 1 dx
Now by part a) it suffices to show that 1 x 3 converges. Indeed,
Z 1
Z 1
1
x 3=2 dx
p dx D
3
x
1
1
ˇ1
h
1=2 ˇ
D 2 x
ˇ
1
D
1
2 lim p D 2:
x!1
x
==
7. Consider the equation
y 0 C 2y D e x :
(9)
a) Find an integrating factor.
Solution. Observe that (9) is of the form
dy
C P .x/y D Q.x/;
dx
where P .x/ D 2 and Q.x/ D e x . Therefore an integrating factor is
I De
R
2 dx
D e 2x :
==
b) Find the general solution.
Solution. Multiplying both sides of (9) by the integrating factor I found in part a) gives
Z
1
2x
ye D e 3x dx
)
ye 2x D e 3x C c:
3
Now
1
y D e x C ce
3
2x
(10)
==
c) Find the solution that satisfies the initial condition y.0/ D 1.
Solution. Let y D 1 and x D 0 in (10), revealing the relation 1 D 1=3 C c. Therefore
c D 2=3 and
1
2
y D e x C e 2x
3
3
is the desired solution.
==
5
8. Write the decimal x D :24
a) as a geometric series
Solution. Note that :24 D :242424 : : : and :24 D 24=100 D 24=102 . Similarly, note that
:0024 D 24=104 and :000024 D 24=106 . Continuing this logic,
2k
1
X
24
24
24
1
:24 D 2 C 4 C 6 C D
24
:
10
10
10
10
nD1
==
b) as a reduced fraction.
Solution. Let x D :24. Then 100x D 24:24. Subtracting 100x from x gives
100x
x D 24
)
99x D 24
So :24 D x D 24=99 D 8=33
==
9. The rhombus whose vertices are . 1; 0/; .0; 3/; .1; 0/; and .0; 3/ is revolved about the line
x D 5. Use Pappas Theorem to compute the volume generated.
Solution. Recall the Theorem of Pappas:
“Let R be a plane region that lies entiredly on one side of a line ` in the plane. If R
is rotated about `, then the volume of the resulting solid is the product of the area A
of R and the distance d traveled by the centroid of R.”
Graphing the rhombus under question, one can see that the figure is symmetric about the x-axis
and the y-axis, implying that the centroid is the origin .0; 0/.
Viewing the rhombus as two triangles, one above the x-axis and one below the x-axis—call
them T1 and T2 respectively, one sees that the area of T1 is .1=2/.2/.3/ D 3 and the area of T2
is .1=2/.3/.2/ D 3. It follows that the area of the rhombus is 3 C 3 D 6.
The perpendicular distance from .0; 0/ to the line x D 5 is clearly 5, so the Theorem of Pappas
can be invoked to say the the volume generated is
V D 6 5 D 30:
6
==