Name:
MAT 21b: Practice Final
Saturday, September 8th, 2012
Write legibly and neatly. Show all your work for full credit. You have 100 minutes to complete this exam.
√
1. (10 points) Find the total area between the graph of the function y = x 4 − x2 and the x-axis on the
interval [−2, 2].
This function has zeroes at x = −2, x = 0, and x = 2. Plugging in a few points, its not so hard to figure
out what the graph of this function looks like, but really, we don’t need to know exactly what it looks like we just need to know on which intervals its positive and negative. Since the square root function is always
positive, this function is negative where x is negative (on (−2, 0)), and positive where x is positive (on (0, 2)).
Hence we can compute the area as:
A=−
Z 0 p
x
4 − x2 dx +
−2
Z 2 p
x
4 − x2 dx
0
Note the negative sign on the first√integral to compensate for the fact that y is negative there. Alternatively,
this function is clearly odd, since 4 − x2 is an even function and x is an odd function. Hence the area under
the x axis and above the funciton from x = −2 to x = 0 will be the same as the area under the function and
above the x axis from x = 0 to x = 2. So we could more simply compute the area as:
A=2
Z 2 p
x
4 − x2 dx
0
I will take the second approach, since its simpler and I am lazy. This integral can be computed via the
u-substitution u = 4 − x2 , so du = −2x dx. Hence:
A=2
Z 2 p
x
4 − x2 dx
0
=−
Z u(2)
u1/2 du
u(0)
2 3/2 0
=− u 3
4
2
= − (0 − 8)
3
16
=
3
Be sure that when making a u-substitution for a definite integral, as in this problem, that you handle the
limits correctly.
2. (10 points) Book problem 6.1 #37 and Book problem 6.2 #5.
Note: I will probably only have one ‘find the volume’ question on the real final. However, I’m not
quite a nice enough guy to tell you which method it will be using, so you should know all of the Disc,
Washer, and Shell method.
Answer to 6.1 #37:
We can use the washer method here. The outer radius is given by the constant function y = 1, and the inner
√
radius is given by the function y = cos x. In addition, both of these functions are even, so we can make use
of that fact to integrate only over the positive x portion of the domain, and just double that volume to find
the total volume:
V =2
√
2
π12 − π cos x dx
Z π/2 0
Z
π/2
dx −
= 2π
0
Z π/2
cos x dx
0
π/2
π
= 2π
− sin x
2
0
π
− (1 − 0)
= 2π
2
2
= π − 2π
!
Answer to 6.2 #5:
For this problem we need to use the method of cylindrical shells. Hence the volume will be given by:
Z √3
V=
2πx
p
x2 + 1 dx.
0
We now make the u-substitution u = x2 + 1, du = 2x dx. After substitution (being careful to take care of the
limits of integration), we obtain:
Z 4
u1/2 du
2 3/2 4
= πu 3
V =π
1
1
2
= π(8 − 1)
3
14
= π
3
3. (10 points)
(a) Compute the definite integral:
Z 1
θ cos πθ dθ .
0
We proceed by integration by parts. Define u and dv as follows:
u=θ
dv = cos πθ dθ
1
v = sin πθ
π
du = dθ
Then we have that:
Z 1
0
1
Z
1
1 1
θ cos πθ dθ = θ sin πθ −
sin πθ dθ
π
π 0
0
1
1
= 2 cos πθ π
0
2
=− 2
π
(b) Compute the indefinite integral:
Z
e2x cos 3x dx.
Again, we use integration by parts. Define u and v as follows:
dv = e2x dx
1
v = e2x
2
u = cos 3x
du = −3 sin 3x dx
Integrating by parts once gives us:
Z
1
3
e2x cos 3x dx = e2x cos 3x +
2
2
Z
e2x sin 3x dx
We now integrate by parts again. Choose u and v as follows:
dv = e2x dx
1
v = e2x
2
u = sin 3x
du = 3 cos 3x dx
Integration by parts a second time gives us that:
Z
3
1
e cos 3x dx = e2x cos 3x +
2
2
2x
1 2x
3
e sin 3x −
2
2
Z
2x
e cos 3x dx
Solving for e2x cos 3x dx, we find that:
R
Z
e2x cos 3x dx +
9
4
1
3
e2x cos 3x dx = e2x cos 3x + e2x sin 3x
2
4
Z
2
3
e2x cos 3x dx = e2x cos 3x + e2x sin 3x +C
13
13
Z
4. (10 points) Compute the indefinite integral:
Z
sec3 x tan3 x dx.
There are probably multiple ways to do this problem, but this is the way that springs to mind for me. If I
take out a sec x tan x, I’m left with all even powers of secant and tangent, which is always a good thing. This
suggests the following strategy for simplification:
Z
3
3
Z
sec2 x tan2 x (sec x tan x dx)
Z
sec2 x sec2 x − 1 (sec x tan x dx)
sec x tan x dx =
=
Now letting u = sec x, we have that du = sec x tan x. Hence after substitution we have:
Z
u2 (u2 − 1) du =
u5 u3
− +C
5
3
After substituting back in for u, we obtain our final answer:
Z
sec3 x tan3 x dx =
sec5 x sec3 x
−
+C
5
3
5. (10 points) Compute the average value of the function:
f (x) =
3x + 1
,
x2 − 2x − 15
from x = 0 to x = 2.
Note: this is really a partial fractions question, but since everything we’ve learned requires integration, I can test you on say, average value, or perhaps arc length, while also testing you on techniques
of integration. You should expect me to do things like this on the final, and not necessarily have each
question fall neatly into the different sections of the book.
Recall that the average value of a function f over the interval x = a to x = b is given by:
Av( f ) =
1
b−a
Z b
f (x) dx
a
Of course, the difficult part of this problem will be computing the partial fraction decomposition of the
given function, so let’s do that now. We can factor the denominator: x2 − 2x − 15 = (x − 5)(x + 2). Since
the denominator has only linear, distinct factors, this decomposition is fairly easy:
3x + 1
B
A
=
+
x2 − 2x − 15 x − 5 x + 3
Cross-multiplying and equating the numerators gives us that:
3x + 1 = A(x + 3) + B(x − 5)
Gathering terms in powers of x gives:
3x + 1 = x(A + B) + (3A − 5B)
Since this must be true for all x, we obtain the system of equations:
3 = A+B
1 = 3A − 5B
This can be solved in your favorite way but all ways should lead to the fact that A = 2 and B = 1. Hence the
average value of the function over x = 0 to x = 2 will be given by:
Z 2
Z 1
3x + 1
1 2
2
1
dx =
+
2 − 0 0 x2 − 2x − 15
2 0 x−5 x+3
2
2
1
= ln |x − 5| + ln |x + 3| ln |x + 3|
2
0
1
= (ln 3 − ln 5) + (ln 5 − ln 3)
2
1 3
= ln
2 5
0
6. (10 points) Compute the indefinite integral:
Z
x4
dx.
(1 − x2 )7/2
Note: You may assume that −1 < x < 1.
This problem requires trigonometric substitution. Let x = sin θ . Then dx = cos θ dθ . Hence:
Z
x4
dx =
(1 − x2 )7/2
Z
sin4 θ
cos θ dθ
(1 − sin2 θ )7/2
Z
sin4 θ
dθ
cos6 θ
=
At this point, as is so often the case, there are probably quite a few ways to go. One particularly easy way is
to sepearte this into some tangents and secants, as follows;
Z
sin4 θ
dθ =
cos6 θ
Z
tan4 θ sec2 θ dθ
At this point, we can make a u-substitution: let u = tan θ . Then du = sec2 θ dθ . After substitution, we
obtain:
Z
u5
+C
5
tan5 θ
=
+C
5
u4 du =
By drawing a reference triangle (hard to do in these PDF documents, so come see me in office hours if you
have trouble with this), we obtain that:
x
tan θ = √
1 − x2
Hence our final answer is:
5
Z
x4
1
x
√
+C
dx =
5
(1 − x2 )7/2
1 − x2
7. (10 points) The intensity L(x) of light x feet beneath the surface of the ocean satisfies the differential
equation:
dL
= −kL.
dx
As a diver, you know from experience that diving to 18 feet in the Caribbean Sea cuts the intensity
in half. You cannot work without artificial light when the intensity falls below one-sixteenth of the
surface value. About how deep can you expect to work without artificial light?
From prior work, we know that the solution to this equation is:
L(x) = L0 e−kx
where L0 is the light intensity at the surface. Note that in this equation, x measures the feet beneath the
surface, so positive x values correspond to feet down, which is the opposite of the usual convention. We can
find k from the fact that at x = 18, L = 21 L0 . Hence:
1
L0 = L0 e−18k
2
Solving for k gives us that:
k=−
To find at what depth L is reduced to
1
16 L0 ,
1
ln 2
ln(1/2) =
.
18
18
we simply solve the equation:
x ln 2
L0
= L0 e− 18
16
This gives:
x=
−18 ln(1/16)
ln 16
log 16
4
= 18
= 18 2 = 18 = 72 feet.
ln 2
ln 2
log2 2
1
8. (10 points) Solve the initial value problem:
√
√ dy
x
= ey+ x ,
dx
y(0) = 1.
We first separate the equation:
1 √
dy
= √ e x dx
y
e
x
Integration gives:
−
Z
1 −1/2 x1/2
x
e dx
2
√
−e−y = 2e x +C
−e
−y
Z
dy = 2
The second integral is of course a u-substitution, but its a fairly easy one so I’ve done it in my head and
left out the details. You’re of course free to do this on a more complicated problem, though if you mess up
partial credit will likely be harder to come by. On a problem where the main point is u-substitution, you
should show all of your steps. But this problem is about solving a separable differential equation, so feel
free to take shortcuts in your integration if you can do it accurately. We now have to solve for y:
√
y = − ln −2e x +C
Since y(0) = 1, we have that:
1 = − ln(−2 +C)
Solving for C, we find that:
1
e
So the particular solution to the given initial value problem is:
√
1
x
y(x) = − ln −2e + 2 +
e
C = 2+
9. (10 points) Find the general solution to the differential equation:
dy 2y
+
= 3x + 4.
dx
x
Solving this differential equation requires the use of an integrating factor. The equation is already in standard form, so it’s easy to identify P(x) = 2x and Q(x) = 3x + 4. To find the integrating factor v(x), we simply
compute:
R 2
x dx
v(x) = e
= e2 ln x
= x2
Hence the general solution to this differential equation is given by:
1
y(x) = 2 (3x + 4)x2 dx
x 1 3 4 4 3
= 2
x + x +C
x 4
3
2
3x
4x C
=
+ + 2
4
3 x
Z
10. (10 points) Does the following integral converge or diverge? If it converges, find its value.
Z 1
x ln x dx.
0
The value of this integrand is undefined at x = 0, and hence this integral is improper. So:
Z 1
Z 1
x ln x dx = lim
x ln x dx.
a&0 a
0
To simplify matters, I’ll begin by computing the indefinite integral of ln x ln x, and I’ll forget about the +C,
since we’re going to use this integral to compute a definite integral anyway. We integrate by parts and let
2
u = ln x, dv = x dx. Hence du = 1x dx and v = x2 . This gives that:
Z
x2
1
x ln x dx = ln x −
2
2
Z
x dx =
x2
x2
ln x −
2
4
Hence:
Z 1
0
"
#
x2
x2 1
x ln x dx = lim
ln x −
a&0
2
4 a
a2
1 a2
−
= lim 0 − ln a −
a&0
2
4 4
The hard part about evaluating this limit is determining the behavior of a2 ln a as a goes to 0. We’ll need to
use L’Hôpital’s rule:
lim = a2 ln a
a&0
= lim
ln a
1
a2
1
a
lim −2
a&0 3
a
a&0
=
= lim −2a2
a&0
=0
Hence:
Z 1
x ln x dx =
0
−1
4
11. (10 points) Does the following integral converge or diverge? You must justify your answer.
Z ∞
1
1
xex
dx.
If you attempt to integrate this directly, you’ll find it rather frustrating. Instead, one should use the Direct
Comparison test. We first note that:
a
Z ∞
Z a
1
−x
−x
−x e dx = lim
e dx = lim −e = lim − e−a − e−1 = .
a→∞ 1
a→∞
a→∞
e
1
1
Hence 1∞ e−x is convergent, so we can use it for the comparison test. Noting that x ≥ 1 on this domain, we
see that:
R
Z ∞
1
1
xex
≤
Z ∞
1
1
ex
dx
Since the integral on the right is convergent, the integral on the left must be, also.
12. (10 points)
(a) Find a parametrization for the circle of radius 2, centered at (2, 2).
We simply rescale and shift the standard parametrization for a circle: t = (2 + 2 cost, 2 + 2 sint), for 0 ≤ t ≤
2π.
(b) Compute the arc length of the curve from part (a).
Note: You are, of course, just computing the perimeter of a circle of radius 2, so it should be easy
for you to check that you got the correct answer.
The arc length is given by:
Z 2π
s
S=
0
=
Z 2π p
dx
dt
2
dy
+
dt
2
dt
4 sin2 t + 4 cos2 t dt
0
Z 2π
=2
dt
0
= 4π.
Which is exactly what we expect, given that the perimeter of a cicle of radius r is given by P = 2πr.