Some Practice Problems for MTH U242
1. Calculate the following integrals
√
Z 4 √x
Z 4
Z u=2
dx
e
e x
1/2
√ dx By substitution: u = x ; du = √ , so
√ dx = 2
(a)
eu du = 2(e2 − e).
x
2 x
x
1
x=1
u=1
Z
1
(b)
cos 3x dx = − sin(3x) + C
3
Z
1
(c)
x cos 3x dx By parts: u = x, dv = cos(3x) dx; du = dx, v = sin 3x.
3
Z
Z
1
x cos 3x dx =
x sin 3x − (1/3) sin 3x dx
3
1
1
x sin 3x + cos 3x + C
=
3
9
Z
1
(d)
x2 cos 3x dx By parts; first: u = x2 , dv = cos 3x; du = 2x dx, v = sin 3x.
3
Z
Z
1
x2 cos 3x dx = x2 sin 3x − (2/3)x sin 3x dx
3
Z
2
x sin 3x
=
− (2/3) x sin 3x dx.
3
Z
x2 cos 3x dx =
=
Let R be the region between the curve y = x2 ,
the y-axis, and the line y = 4. Find the volume
obtained when R is rotated about the axes below.
We use the formula:
Z b
Z d
V =
A(x) dx or V =
A(y) dy
a
R
x
1
x sin 3x dx = − cos 3x + sin 3x, so
3
9
µ
¶
x2 sin 3x
x
1
− (2/3) − cos 3x + sin 3x
3
3
9
2
2
x sin 3x 2x
+
cos 3x −
sin 3x + C.
3
9
27
Using integration by parts on this last integral gives
y
4
3
R
2
c
In each case, A = πR2 if there is no hole (crosssections are disks) or A = π(R2 − r2 ) if there is a
hole (cross-sections are washers).
(x,y)
1
0
0
1
2
x
(a) Around the y-axis: There is only one radius since the region goes right up to the axis of rotation.
We slice perpendicular to the axis of rotation, so our slices here have thickness dy:
√
R = x= y
¯4
Z 4
y 2 ¯¯
y dy = π ¯ = 8π.
V = π
2 0
0
(b) Around the x-axis: Since there is a gap between the region and the x-axis, there are two radii. Since
the slices are vertical, their thickness is dx, and we have
r
V
= yinner = x2 , R = youter = 4
Z 2
= π
42 − (x2 )2 dx
0
µ
¶¯2
x5 ¯¯
128π
= π 16x −
=
.
5 ¯0
5
(c) Around the line x = −3: As in (a), the slices are vertical, so have thickness dy, and
r = 3 + xinner = 3
√
R = 3 + xouter = 3 + y
Z 4
√ 2
V = π
(3 + y) − 9 dy
0
= π
µ
¶¯4
¯
y2
+ 4y 3/2 ¯¯ = 40π
2
0
2. A 100 m rope with density 0.5 kg/m has a 50 kg mass attached at the bottom. How much work is done
in pulling up the entire rope and weight?
Suppose x meters of the rope is hanging. It has a mass of x/2 kg, so weighs about 9.8x/2 = 4.9x newtons.
The mass at the bottom adds (9.8)(50) = 490 newtons. Raising this dx meters does (4.9x + 490) dx
joules of work. Thus, the total work done in raising the entire rope is approximately
100
Z
(4.9x + 490) dx =
0
¯100
¯
4.9x2
+ 490x¯¯
2
0
= 73, 500 joules.
3. Does the series
∞
X
(−1)n
n=0
2
10n2
n2
converge? Explain your answer.
− 3n + 2
n
1
n2 /n2
=
=
. Thus, the nth term of the series does not
10n2 − 3n + 2
10n2 /n2 − 3n/n2 + 2/n2
10
approach 0, so the series diverges by either the divergence test or the alternating series test.
limn→∞
4. What does
∞
X
n=0
∞
X
7n
32n+1
converge to? (Write out a few terms.)
∞
X
7n
1
1
7
49
. This is a geometric series with first term and common
+
+
+
·
·
·
=
2n+1
n
3
3
27
243
3
·
9
3
n=0
n=0
7
a
1/3
3
3
ratio . Thus, its limit is
=
=
= .
9
1−r
1 − 7/9
9−7
2
7n
=
5. Find the solution to each of the following initial-value problems.
(a) x3
dy
= 2y, y(1) = 4
dx
Separate variables and integrate:
1
dy
y
Z
1
dy
y
ln y
y
2
dx
x3
Z
Z
2
dx
=
2
x−3 dx
=
x3
= −x−2 + C = −1/x2 + C
=
= Ke−1/x
2
2
2
Letting x = 1, y = 4: 4 = Ke−1 , K = 4e, so y = 4e · e−1/x = 4e1−1/x .
(b)
t2 + 1
dx
=
, x(0) = 2.
dt
x
Separate variables and integrate:
x dx = t2 + 1 dt
x2
t3
=
+t+C
2
3
2t3
+ 2t + 2C.
x2 =
3
p
2t3 /3 + 2t + 4. Since x(0) = 2 > 0, we
Letting t = 0, x = 2 gives 4 =
2C,
C
=
2.
Thus,
x
=
±
p
3
choose the positive root: x = 2t /3 + 2t + 4.
6. A vertical cylindrical tank has radius 2 meters and height 5 meters. Its top is 1 meter below ground level,
and it is 80% filled with water (density = 1, 000 kg/cu.m).
(a) Draw a diagram of the situation.
1m. = below ground
2 m. = radius
5 m. = tank height
h m. = water level
(b) Set up an integral to compute the work done in pumping all the water in this tank to ground level.
Show in your diagram what your “height” variable represents.
We calculate the work necessary to raise a thin “film” of water to ground level. First, if the film is a
cylinder of thickness dh, then its volume is πr2 dh = 4π dh. Since the density is 1000, this has mass
4000π dh (kg), hence weight (9.8)4000π dh. As you can see from the diagram, this must be lifted
a distance of 5 − h to reach the top of the tank, and another 1 meter to ground level, so total lift
is 6 − h. The work done on this volume of water is then (9.8)4000π(6 − h) dh joules, so the total
work–since the tank is 80% full–is
Z h=4
(9.8)4000π
(6 − h) dh.
h=0
(c) (Optional) Calculate the integral.
(d) Do a similar problem, but with a conical tank of height 5 meters and radius 2 meters (at the bottom).
Assume it is completely filled at the beginning.
Here is a revised diagram:
1m.
y
5 m.
r
2 m.
r
y
2y
From similar triangles,
= , so r =
, and the volume of the thin film is approximately
2
5
5
µ 2¶
4y
dy. As before, we multiply by (9.8)1000 to get the weight (force). The lift is
πr2 dy = π
25
now y + 1, so the work in joules is given by
Z 5
Z
4y 2
9.8(4000)π 5 3
(y + y 2 ) dy
(9.8)1000π
(y + 1) dy =
25
y=0 25
0
7. Find the area between the curves y + 4 = x2 and y − 2x = 4.
y
10
7.5
5
2.5
0
-4
-2
0
2
4
x
-2.5
¯4
R4
(2x + 4) − (x2 − 4) dx = −2 2x − x2 + 8 dx = x2 − x3 /3 + 8x¯−2 = 36.
R3
8. Use Simpson’s rule with n = 6 to approximate 2 sint t dt. (Answer to 4 places.)
Area =
R
(top − bottom) dx =
R4
−2
t
2
13/6
14/6
sin t
.4546 .3820 .3099
t
weight
1
4
2
terms: .4546 1.5280 .6198
= .2432. (Exact = .2432391627).
15/6
16/6
17/6
3
.2394
.1715
.1071
.0470
4
.9576
2
3430
4
.4283
; Sum = 4.3783;
1
.0470
R3
2
sin t
t
dt ≈
1/6
Sum
3
9. Which of these improper integrals converge?
R ∞ dx
M
(a) 0 1+x
= limM→∞ arctan(M ) = π/2. This converges.
2 = limM →∞ arctan(x)|0
R 2 dx
2
(b) 1 x−1 = lima→1 ln(x − 1)|a = lima→1 ln 2 − ln(a − 1). This diverges since limu→0 ln u = −∞.
10. What is the interval of convergence of
We use the ratio test:
P∞
k
k (x+1)
k=1 (−1)
k2k
?
¯
¯
¯ (x + 1)k+1
k2k ¯¯
|x + 1|
k
|x + 1|
¯
= lim
·
lim ¯
=
·
.
¯
k+1
k
k→∞ (k + 1)2
k→∞
(x + 1)
2
k+1
2
| {z }
approaches 1
This will be < 1 when |x + 1| <
2
|{z}
radius of conv.
; i.e. when −2 < x + 1 < 2, or −3 < x < 1. Now we test the
P∞
P∞ 1
(−2)k
endpoints. Putting x = −3 gives the series k=1 (−1)k
= k=1 ; this is the harmonic series, so
k2k
k
it diverges. Putting in x = 1 gives the alternating harmonic series, which converges. Thus, the interval
of convergence is precisely −3 < x ≤ 1.
11. Below are formulas for the derivatives of two functions P and Q. In each case, use the formula to find
the coefficient of the kth degree term in the appropriate Taylor series expansion for the function, then
write the first 4 terms of this expansion. Find the radius of convergence for the Taylor series.
∞
X
f (k) (a)
[We recall that the Taylor series for f is T (f, a)(x) =
(x − a)k , where f (k) denotes the kth
k!
k=0
derivative.]
k4k
3k+1
∞
X
k4k
4
192
32
T (P, 5)(x) =
(x − 5)k = 0 + (x − 5) + (x − 5)2 +
(x − 5)3 + · · ·
k+1
3 k!
9
54
486
k=0
¯
¯
¯ (k + 1)4k+1 (x − 5)k+1
4 |x − 5|
3k+1 k! ¯¯
k+1
4 |x − 5|
¯
=
The ratio test gives ¯
· k
·
=
−→
¯
k+2
k
3 (k + 1)!
k4 (x − 5)
3
k(k + 1)
3k
0 < 1, so this converges for all x; i.e., the radius of convergence is infinite.
(k + 1)!4k
(b) Q(k) (2) =
3k+1
∞
∞
X
X
1 8
256
16
(k + 1)!4k
(k + 1)4k
k
T (Q, 2) =
=
(x−2)k = + (x−2)+ (x−2)2 +
(x−2)
(x−2)3 +· · ·
k+1
k+1
3 k!
3
3 9
9
81
k=0
k=0
¯
¯
¯
¯ (k + 2)4k+1 (x − 2)k+1
3k+1
¯ = 4 |x − 2| · k + 2 −→ 4 |x − 2| .
¯
·
The ratio test gives ¯
3k+2
(k + 1)4k (x − 2)k ¯
3
k+1
3
3
This will be less than 1 when |x − 2| < , so the radius of convergence is 3/4.
4
(a) P (k) (5) =
12. Find
R
arctan x dx
R
1
dx and v = x: arctan x dx = x arctan x −
Integration by parts: u = arctan x, dv = x, so du =
2
1+x
Z
1
x
2
= arctan x − ln(1 + x ) + C (the last integral is by substitution: u = 1 + x2 ).
1 + x2
2
1
, find the power series for arctan x, then the power series for
1 + x2
Z 0.2
arctan x
arctan x
. Use this last power series to estimate
dx with an error of less than 0.0005.
x
x
0
(Explain how you know that you have this accuracy.)
1
= 1 − u + u2 − u3 + · · · , let u = x2 :
Starting from
1+u
1
= 1 − x2 + x4 − x6 + · · · ; now integrate:
1 + x2
x3 x5 x7
arctan(x) = x −
+
−
+ ···
3
5
7
x2 x4
x6
arctan x
= 1−
+
−
+ ···
x
3
5
7
¯0.2
Z 0.2
¯
arctan x
x3 x5 x7
dx = x − 2 + 2 − 2 + · · · ¯¯
x
3
5
7
13. Starting with the power series for
0
0
0.23 0.25 0.27
= 0.2 − 2 + 5 − 2 + · · ·
3
5
7
0.22K+1
. Checking
(2K + 1)2
0.23
0.25
with a calculator gives
≈ 0.00089 which is too big, so the earliest term we can drop off is 2 . Thus
9
5
R 0.2 arctan x
0.23
dx ≈ 0.2 − 2 ≈ 0.19911.
0
x
3
14. Test these series for convergence:
Since this is an alternating series, the truncation error is < the first term dropped =
∞
X
∞
X 2n
2n
. The last series is a convergent geometric series, so the first series converges
<
3n + 5
3n
n=0
n=0
by the comparison test.
¯
¯
∞
X
¯
¯ (n + 2)35n+1
n!
(n + 1)35n
¯ = (n + 2)35 −→ 0, so this
(b)
; we use the ratio test: ¯¯
·
¯
n
n!
(n
+
1)!
(n
+
1)35
(n + 1)2 n→∞
n=0
converges.
∞
X
(−1)n+1
1
√
(c)
; this is an alternating series and √ decreases to 0; thus, it converges by the alternating
n
n
n=1
series test.
∞
∞
X
X
6
1
√
=
6
. This is a p-series with p = 3/4 < 1, so it diverges by the p-series test. (You
(d)
4
3/4
3
n
n
n=1
n=1
can also use the integral test.)
(a)
15. An exercise spring at a gym requires a force of 20 Lbs to stretch it from its natural length of 2 feet to a
length of 3 feet. How much work (foot-pounds) is done in stretching it to a length of 4 feet, 25 times?
Hooke’s Law says F(x) = kx where x is the stretch beyond natural length. Thus, 20 = k · 1, so k = 20
for this spring. We now can compute the work done in stretching it 2 feet, 25 times:
Z 2
W = 25
20x dx = 1000 (foot-pounds).
0
16. A particle moves according to the parametric equations x = 3 cos t, y = 5 sin t. Eliminate the parameter
t to get a simple equation with just x and y (HINT: look at x/3 and y/5). If t goes from 0 to π, describe
the path of the particle. Set up, but do not evaluate, an integral to find the distance it travels.
x/3 = cos t and y/5 = sin t.
x
y2
+
= cos2 t + sin2 t = 1,
33
52
2
so this is an ellipse. As t goes from 0 to π, the point moves from (3, 0) through (0, 5) to (−3, 0) – i.e.
counter-clockwise through 1/2 of the ellipse.
Z π p
Z πp
s=
(x0 )2 + (y 0 )2 dt =
9 sin2 t + 25 cos2 t dt.
t=0
0
17. The force F = h2, 3, 3i moves an object along the line h3 + t, 5 + 2t, 4 − ti at t goes from 0 to 3. How
much work is done?
Work = Force · Displacement.
As t goes from 0 to 3, the object moves in a straight line from (3, 5, 4) to (6, 11, 1), so the displacement
is h3, 6, −3i; the work is then h2, 3, 3i · h3, 6, −3i = 15.
18. Set up, but do not evaluate, an integral to compute the arc length of the curve y = sin 2x, from x = 2 to
x = 3.
Z 3 p
Z 3 p
0
2
s=
1 + (f (x)) dx =
1 + 4 cos2 2x dx.
x=2
x=2
19. Find the radius of convergence and the actual interval of convergence (test the endpoints) for the power
∞
X
(−1)n+1 4n (x − 2)n
.
series
n3n
n=0
We use the ratio test:
¯
¯ n+1
¯ 4n |x − 2|
¯4
n3n
(x − 2)n+1
4
¯=
¯
·
−→ |x − 2| .
¯ (n + 1)3n+1
n
n
4 (x − 2) ¯
3(n + 1) n→∞ 3
3
3
3
5
11
5
This must be < 1, so we get |x − 2| < , so − < x − 2 < , i.e.
<x<
. When x = the series
4
4
4
4
4
4
becomes
µ
µ
¶
¶
∞
∞
n
n
X
X1
4
3
1
(−1)n+1
−
=−
= −∞ (diverges).
3
4
n
n
n=0
n=0
11
the series is the alternating harmonic series, so it converges. Thus, the interval of
4
5
11
convergence is < x ≤
.
4
4
R
20. Calculate x10 ln x dx.
When x =
Integration by parts: u = ln x, dv = x10 dx; du = (1/x) dx, v = (x11 /11).
Z
Z
10
11
x ln x dx = x ln x/11 − (1/11) x10 dx
=
x11 ln x x11
−
+ C.
11
121
21. Let A = (2, 1, 3) and B = (1, 0, 4). Find the vector projection ProjB A.
µ
¶
A·B
14
B = (1, 0, 4).
ProjB A =
B·B
17
22. Convert from polar to rectangular: θ = 2.91, r = 7. Convert from rectangular to polar: (−5, −12).
(x, y) = (7 cos(2.91), 7 sin(2.91)) ≈ (−6.81, 1.61).
p
√
r = x2 + y 2 = 25 + 144 = 13; tan θ = 12/5 and θ is in quadrant III, so θ = π + arctan(12/5) ≈ 4.32.
23. Sketch the curve r(t) = (3 cos(t), 3 sin(t), 4t). Find its unit tangent T(t). Write an integral that expresses
its arclength, measured from r(0).
This is a helix of radius 3, along the z-axis.
r0 (t) = (−3 sin t, 3 cos t, 4)
p
kr0 (t)k =
9 sin2 t + 9 cos2 t + 16 = 5
µ
¶
r0 (t)
3
3
4
T(t) =
= − sin t, cos t,
kr0 (t)k
5
5
5
Z t
Z t
kr0 (u)k du =
5 du = 5t.
s(t) =
u=0
u=0
24. Find the parametric equations of the line joining (2, 2, 13) and (1, −1, 5).
x(t) = (1 − t)P + tQ
= P + t(Q − P)
= (2, 2, 13) + t(−1, −3, −8) = (2 − t, 2 − 3t, 13 − 8t).

x=2−t 
y = 2 − 3t

z = 13 − 8t