Final exam review sheet answers, Spring 2009
Problem 1.
x + 1F
SinB
á
âx
x
Set u =
1
x +1, du =
2
dx.
x
xE
-2 CosA1 +
2x
à Ix ã M â x
Use integration by parts with u = x, dv = e2 x dx.
E2 x -
1
+
4
à
x
2
x
âx
x2 + 3 x + 2
Use integration by partial fractions.
-Log@1 + xD + 2 Log@2 + xD
à
А2
0
Cos@tD
ât
3 Sin@tD + 1
Use substitution u = 3 Sin[t] + 1, and update the limits of integration.
2
3
á
1
4+
âx
x2
2
final review answers.nb
Use the trig substitution x = 2 Tan[t], dx = 2 Sec2 [t]dt, to obtain LogB
4+x2
2
+
x
F,
2
which is the same as ArcSinh[x/2]
as reported by Mathematica.
3
à Cos@xD â x
Split off one copy of Cos[x] to go with the dx, rewrite the Cos2 [x] as 1 - Sin2 [x], and set u = Sin[x], du = Cos[x]dx.
Sin@xD - Sin@xD ^ 3  3 + C
-2 x
â x = limt®¥ à ã-2 x â x = limt®¥
à ã
¥
t
1
1
â
¥
1
1
2
ã2
-
1
ã2 t
=
1
2 ã2
n
2
5
n=2
A converging geometric series with a = 4/25 and r = 2/5.
4
15
Problem 2.
à
¥
x
1 x3
âx
+ 10
The integrand "resembles"
clearly less than
1
x2
1
x3
<
1
x2
for large x, so we suspect the improper integral converges. Indeed, the integrand is
for all x >= 1, and therefore our integral converges by comparison to the known converging integral Ù1
¥ 1
x2
â x.
We can make our tail less than a given amount by making the corresponding tail of the "test function" that small. So, to
find d as requested, we compute
à
¥
d
1
x2
âx =
1
d
and solve the inequality d1 < .001, to obtain d > 1000. Any value of d larger than 1000 will do.
Let's test this:
final review answers.nb
In[4]:=
Out[4]=
NBà
¥
3
1
â xF
1001
x3 + 10
4.99001 ´ 10-7
We see that this "tail" is considerably less than .001.
Problem 3.
The slices of the x-axis volume are disks of radius ãx . The slices of the y-axis volume are disks of radius 1 between y = 0
and y = 1, and are washers of inner radius ln[y] and outer radius 1 for 1 <= y <= e. (The y-axis volume is easier to compute using
the "shell method.")
xAxisVolume = Π à ã2 x â x
1
0
-
1
+
2
E2
Π
2
yAxisVolume = Π + Π à I12 - Log@yD2 M â y
ã
1
2Π
Problem 4.
If f[x] =
Ù1
3
1+
9
x4
3
,
x
then f'[x] =
-3
x2
. The length of the graph of f on the interval [1,3] is therefore Ùa
b
1 + f '@xD2 â x=
â x. The length can be estimated by adding up the distances between the points (1,3), (2, 32 ), and (3,1):
H2 - 1L2 +
approxArcLength = NB
3
2
2
-3
+
H3 - 2L2 + 1 -
3
2
2
F
2.92081
The Right Riemann Sum R2 for the arc length integral is (delta-x = 1, and the right endpoints of the two subintervals are x = 2
and x = 3):
NB
H3 - 1L
2
1+
9
H2L
4
+
1+
9
34
F
2.30409
The Right Riemann Sum gives an underestimate of the arc length (as does the polygonal approximation), since the integrand is
decreasing.
4
final review answers.nb
The Right Riemann Sum gives an underestimate of the arc length (as does the polygonal approximation), since the integrand is
decreasing.
Problem 5.
The problem is to find the number of subintervals needed to estimate the value of the (easy) integral Ù0 I4 - x2 M â x to
2
within .001 using the Midpoint method. The error bound for the Midpoint method is given by
tive of the integrand is the constant -2, so K2 = 2. We must solve the inequality
2 H2-0L3
24 n2
K2 Hb-aL3
24 n2
. Now the second deriva-
< .001, or n2 >
16 000
,
24
or n >
16 000
24
=
25.8199. Therefore n = 26 will do.
Problem 6.
1/3, 2/5, 3/7, 4/9, 5/11, ..., n/(2n+1), ...
1, -1/2, 1/6, -1/24, 1/120, ..., H-1Ln+1 /n!, ...
Problem 7.
{2 - 1/n} converges monotonically to 2: 1, 3/2, 5/3, 7/4, ...
-1, -2, -3, -4, -5, ... or -1, -4, -9, -16, -25, ... diverge to -¥.
Ú¥
k=1
1
1
is a diverging p-series.
k2
Ú¥
k=1 I
-1 k
M
3
is a converging alternating geometric series.
Problem 8.
a. diverges by comparison to diverging p-series with p = 2/3.
b. converges (use the ratio test).
c. diverges (the nth term tends to ¥ as n ® ¥).
d. converges by the alternating series test.
Problem 9.
The given power series is geometric, with common ratio given by
x-2
.
2
The series will converge for all x such that | x-2
|<
2
1, that is, |x-2| < 2. In other words, the series converges for all x that lie less than 2 units from the center 2, which yields the open
interval (0, 4).
final review answers.nb
Problem 10.
The slope y' is twice the y-coordinate, so the curve is a solution of the DE y' = 2y, which has solutions of the form y = A*ã2 x . If
y = 5 when x = 0, the constant A must be 5. If the slope is twice the x-coordinate, then we have the DE y' = 2x, or y =
x2 + C, and C = 5 to satisfy the initial condition.
Problem 11.
The number of bacteria at time t is given by PHtL = P0 ãk t , where t is in hours. We are given that PH2L = 600 and PH8L = 75 000.
PH8L
PH2L
Therefore
=
P0 ã8 t
P0 ã2 t
=
75 000
600
PH2L = 600 = P0 ã.804719×2 Þ P0 =
= 125 Þ ã6 k = 125 Þ k =
600
ã2 k
ln@125D
6
=
.804719.
Solving
for
P0
in
= 120. Therefore, PHtL = 120 ã.804719 t . The population at 5 hours is PH5L = 6708. The
growth rate at 5 hours is P¢ HtL = 120 kãk t = 5398 bacteria/hour. To find when the population will reach 200,000 bacteria, we
solve the equation PHtL = 200, 000 for t. The answer is: t = 9.21885 hours.
Problem 12.
The graph of r = sinH2 ΘL is the four-leafed rose shown. The area swept out on the interval 0 £ Θ £ Π  2 is the area of the leaf in
the first quadrant. This is the value of the following integral:
à
А2
0
1
Sin@2 tD ^ 2 â t
2
Π
8
ParametricPlot@8Sin@2 tD Cos@tD, Sin@2 tD Sin@tD<,
8t, 0, 2 Π<, AspectRatio ® AutomaticD
0.75
0.5
0.25
-0.75 -0.5 -0.25
-0.25
-0.5
-0.75
0.25
0.5
0.75
5
6
final review answers.nb
Problem 13.
The slope at a point on the parametric curve HxHtL, yHtLL is given by
y¢ HtL
x¢ HtL
. In this case, the slope is
3 t2 -12
.
-2 t
The slope is horizontal
when the numerator is zero and the denominator isn't 0; these values are t = ±2, corresponding to the points H6, -16L and H6, 16L.
The slope is vertical when the denominator is 0 and the numerator isn't; the only such point is t = 0, corresponding to the point
H10, 0L. Here is (a portion of) the curve:
pcurve = ParametricPlotA910 - t2 , t3 - 12 t=, 8t, -3, 3<E
15
10
5
2
4
6
8
10
-5
-10
-15
The slope at t = -1 is
-9
,
2
and the point is H9, 11L, so an equation of the line is Hy - 11L =
lplot = PlotB11 -
9 Hx - 9L
2
, 8x, 6, 12<F
Show@lplot, pcurveD
20
10
2
-10
4
6
8
10
12
-9
2
Hx - 9L.
final review answers.nb
7
Problem 14.
To find the radius of convergence of a power series, one uses the ratio test to check where the series converges absolutely.
Then, once the radius of convergence R is known, one needs to check the endpoints c-R, c+R (where c is the center of the power
series) separately for convergence.
a. center = 1, R = 4, series converges at x = -3, diverges at x = 5, so the interval of convergence is [-3, 5).
b. center = 1, R = 1, interval of convergence [0, 2).
c. center = -5, R = 1, interval of convergence [-6, -4].
d. center = 1, R = 12 , interval of convergence [1/2, 3/2).