Name:
Math 1220
Final Exam Review
This review is a fairly comprehensive picture of what we have covered this semester, and all1 of these
problems are of an appropriate level2 for the final exam.
1. Compute the following derivatives:
(a) Dx ln(sin(x))
Dx ln sin x =
cos x
sin x
(b) Dx 2cos(x)
−2cos(x) log (2) sin (x)
(c) Dx x4x
We need to use logarithmic differentiation here:
y = x4x
ln y = ln x4x
ln y = 4x ln x
Dx ln y = 4 ln x +
4x
x
y0
= 4 ln x + 4
y
y 0 = 4(ln x + 1)y
y 0 = 4(ln x + 1)x4x
2. Compute the following integrals. Note that I have deliberately not told you which techniques to use; if
you are pressed for time, you can still study in a useful way by simply figuring out which technique to
use for each, without computing the integral.
Z
(a)
axn dx
a
Z
(b)
xn+1
+ C if n , 0; a ln |x|+C if n = −1.
n+1
2
ex dy
2
Notice that the integral is dy so the answer is yex + C.
1 The last two problems are actually quite challenging, since you have not seen many problems of that sort before, so they would
be considered challenge problems on the exam as well.
2 That’s not to say that they are all the same difficulty level, of course.
Final Exam Review
Z
cos x
(c)
dx
1 + sin x
Math 1220
ln |1 + sin x| + C; try u = sin x.
Z
(d)
ex sin ex dx
− cos(ex ) + C; try u = ex .
Z
1
dx
x ln(3x)
(e)
ln | ln(3x)| + C; try u =
1
.
3x
Z 2√
(f)
4 − x2 dx
−2
2π; Note that the integral represents the area of the upper half disk {(x, y)|x2 + y 2 ≤ 4, y ≥ 0}.
Z
1
dx
1 − 4x2
(g)
1
1 + 2x
1
1
1
1
ln |
| + C; we need partial fraction decomposition
= (
+
).
2
4
1 − 2x
2 1 − 2x 1 + 2x
1 − 4x
Z
1
(h)
0
2x
dx
1 + x4
R 2x
π
dx = arctan(x2 ) using u = x2 .
; first show that
4
1 + x4
4ex
dx
√
1 − e2x
Z
(i)
4 arcsin(ex ); try u = ex .
π
6
Z
(j)
(sin x)2cos x dx
0
√
R
1
1 cos x
(2 − 2 3/2 ); (sin x)2cos x dx = −
2
.
ln 2
ln 2
Z
(k)
tan x dx
− ln | cos x| + C, recall tan x = sin x/ cos x,try u = cos x.
Z
(l)
xex dx
(x − 1)ex ; integral by parts u = x, v = ex , udv = xex dx.
Final Exam Review
Z
(m)
ln x dx
Math 1220
(ln |x|−1)x, integral by parts, u = ln x, v = x; or let x = et then ln xdx = tet dt and apply the result
of the previous question.
Z √
(n)
x3 1 − x2 dx
√
3
1
(1 − x2 ) 2 (3x2 + 2) + C; rationalizing the integrand by substituting u = 1 − x2 . Trig substi15
tution u = sin x, x ∈ [−π/2, π/2] works similarly.
−
π
2
Z
(o)
sin6 x dx
0
5
1
3
1
x− sin(2x)+
sin(4x)+
sin3 (2x)+C; Use half(double) angle formula to simplify the in16
4
64
48
1
1
tegrand, sin6 x = [(1 − cos 2x)/2]3 = [1 − 3 cos 2x + 3 cos2 2x − cos3 2x], cos2 2x = (1 + cos 4x).
8
2
Z
(p)
tan6 x dx
−x + tan x −
1
1
tan3 x + tan5 x + C ; recall tan2 x = sec2 x − 1 = (tan x)0 − 1, so
3
5
tan6 x = tan4 x((tan x)0 − 1) = tan4 (tan x)0 − tan4 x
= tan4 (tan x)0 − [tan2 x((tan x)0 − 1)] = tan4 (tan x)0 − tan2 x(tan x)0 + tan2 x
= tan4 (tan x)0 − tan2 x(tan x)0 + (tan x)0 − 1.
Z
(q)
sin2 3x cos2 3x dx
x sin(12x)
−
+ C; apply the half angel formulas to the integrand.
8
96
Z
(r)
x6
dx
(x − 2)2 (1 − x)5
72
15
64
8
1
+
+
+
+
+ 128 ln |x − 2| − 129 ln |x − 1| + C; check your partial
x − 2 x − 1 (x − 1)2 3(x − 1)3 4(x − 1)4
fractions with WolframAlpha by entering, for example, " partial fractions x6 /((x − 2)2 (1 − x)5 )”.
Z
(s)
2x2 + x − 4
dx
x3 − x2 − 2x
2 ln |x| − ln |x + 1| + ln |x − 2| + C, partial fractions
Z
(t)
cos x
sin4 x − 16
dx
2x2 + x − 4
2
1
1
= −
+
.
3
2
x − x − 2x x x + 1 x − 2
Final Exam Review
Math 1220
R
1 2 − sin x
sin x 1
1
2−u
ln(
) − 2 arctan(
) ; Try u = sin x, antiderivative of
du =
(ln |
|−
4
32
2 + sin x
2
32
2+u
u − 16
u
2 arctan( )) + C.
2
Z∞
1
(u)
dx
1.000001
x
0
Divergent; we have infinite integrand at x = 0 and infinite limit,
R1
R∞
1
1
dx
diverges,
dx converges.
0 x1.000001
1 x1.000001
Z
∞
(v)
−∞
1
x2 + 2x + 10
dx
π
1
x+1
π
; infinite limits,antiderivative arctan(
) + C, limt→∞ arctan(t) = , limt→−∞ arctan(t) =
3
3
3
2
π
− .
2
Z∞
2
(w)
xe−x dx
1
1
1 2
; Infinite limit, antiderivative − e−x + C.
2e
2
Z
(x)
3
1
dx
3
−1 x
Divergent. Infinite integrand at x = 0, antiderivative −
Z
3
(y)
√
x
9 − x2
0
1
1
+ C, limx→0 2 = ∞.
x2
x
dx
√
3; Infinite integrand at x = 3, antiderivative − 9 − x2 + C.
π
2
Z
(z)
0
cos x
dx
1 − sin x
Divergent. Infinite integrand at x = π/2, antiderivative − ln |1 − sin x| + C.
3. Compute the following limits.
(a) limx→0+
x2
sin x−x
lim+
x→0
0/0
x2
2x
= lim+
sin x − x x→0 cos x − 1
0/0
2
= lim+
x→0 − sin x
= −∞
Final Exam Review
(b) limx→0+
Math 1220
cos x
x
Don’t use L’Hôpital’s rule here. Since cos(0) = 1 the answer is simply ∞.
1
(c) limx→∞ x x
This is of the form ∞0 , so we need to use a natural logarithm:
1
1
ln lim x x = lim ln x x
x→∞
x→∞
1
ln(x)
x
ln(x)
= lim
x→∞ x
= lim
x→∞
∞
∞
=
lim
x→∞
1
x
1
=0
But that’s not the answer. This is:
1
x
lim x = e
1
ln limx→∞ x x
x→∞
= e0
=1
3
(d) limx→∞ xex
x3 ∞
3x2
∞
=
lim
x→∞ ex
x→∞ ex
∞
6x
∞
=
lim
x→∞ ex
∞
6
∞
=
lim
x→∞ ex
=0
lim
(e) limx→0+ x ln x
Final Exam Review
Math 1220
This is of the form 0 · ∞, so:
ln x
x→0+ 1/x
∞
1/x
∞
=
lim
x→0+ −1/x2
= lim −x
lim x ln x = lim
x→0+
x→0+
=0
(f) limn→∞ sinn n
First let’s deal with that sin n. First note that:
−1 ≤ sin n ≤ 1
1 sin n 1
− ≤
≤
n
n
n
So by the squeeze theorem:
lim −
n→∞
sin n
1
1
≤ lim
≤ lim
n→∞ n
n n→∞ n
sin n
0 ≤ lim
≤0
n→∞ n
So that means that limn→∞ sinn n = 0.
4. Determine if each of the following series converge or diverge.
P
1
(a) ∞
k=2 k ln k
This is easiest with the integral test. Note that f (x) =
1
1). Also, since x ln x is increasing, x ln
x is decreasing. So:
∞
Z
2
1
dx = lim
x ln x
b→∞
Z
b
2
1
x ln x
is continuous and positive (for x >
1
dx
x ln x
Now use the substitution u = ln x:
Z
= lim
b→∞
= lim
b→∞
ln b
1
dx
u
ln 2
ln b
ln uln
2
= lim [ln ln b − ln ln 2]
b→∞
=∞
Since the integral diverges, so does the sum.
(b)
−k 2
k=1 k 10
P∞
Final Exam Review
Math 1220
2
Again, let’s just use the integral test. The function f (x) = x10−x is a positive, continuous function. Let’s see that it’s decreasing:
2
2
2
d x10−x = 10−x − 2x−1 x ln(10)10−x
dx
2
= (1 − 2 ln(10))10−x
2
So since 10−x is always positive, this is always negative, and f (x) is always decreasing. Now:
Z
∞
x10
−x2
b
Z
dx = lim
b→∞
1
2
x10−x dx
1
b
2
1
1
− 10−x
2
b→∞ ln 10
1
2 ib
1 h −b2
10 − 10−1
= lim −
1
b→∞ 2 ln 10
ib
2
1 h
=−
−10−1
1
2 ln 10
1
=
20 ln 10
= lim
Since the integral converges, so does the sum.
(c)
P∞
n
n=1 n2 +2n+1
Let’s use the Limit Comparison Test, with
series):
P∞
1
n=1 n
an
= lim
n→∞ bn
n→∞
lim
= lim
n→∞
(which diverges, since it’s the harmonic
n
n2 +2n+1
1
n
n2
n2 + 2n + 1
1
n2
1
n2
1
n→∞ 1 + + 1
n2
= lim
2
n
=1
(d)
7n
n=1 n!
P∞
Ratio test:
7n+1 n!
an+1
= lim
n→∞ an
n→∞ (n + 1)! 7n
7
= lim
n→∞ n + 1
=0
lim
So the series converges.
(e)
P∞
sin n
n=1 n2
Final Exam Review
Math 1220
This is not a positive series. Convergence means either absolutePor conditional
convergence.
sin n instead.
Let’s shoot for absolute convergence by testing the convergence of ∞
n=1 n2
Notice that
sin n = | sin n| ≤ 1
n2 n2
n2
P∞ sin n P∞ sin n
P
1
Since ∞
n=1 n2 converges (since it’s a p-series), the series n=1 n2 converges, and so n=1 n2
converges.
(f)
n
1
n=1 n 2
P∞
P
n
First of all, let’s simplify this to ∞
n=1 2n . It’s not really obvious how to proceed here, but the
best bet is probably the ratio test, since there is an n in an exponent:
n + 1 2n
an+1
= lim n+1
n→∞ 2
n→∞ an
n
n+1 1
= lim
n→∞ n 2
1
n+1
= lim
2 n→∞ n
1
1
= lim 1 +
2 n→∞
n
1
= <1
2
lim
So the sum converges.
5. Determine if the following series converge or diverge. If they converge, evaluate the sum (you may
have to be very clever).
P
k −k
(a) ∞
k=1 3 5
∞
X
3k 5−k =
k=1
=
=
∞
X
3k
5k
k=1
∞ X
k=1
3
5
1 − 35
3
5
= 2
5
=
(b)
P∞
1
k=2 k 2 −1
3
5
3
2
k
Final Exam Review
Math 1220
Using partial fractions, you can come up with the following equality:
n 1X 1
1
1
=
−
k−1 k+1
k2 − 1 2
k=2
k=2
1 1
1
1
1
1
1
=
−
+
−
+
−
+
2 2−1 2+1 3−1 3+1 4−1 4+1
1
1
1
1
1
1
··· +
−
+
−
+
−
n−2−1 n−2+1 n−1−1 n−1+1 n−1 n+1
1 1 1 1 1 1 1
1
1
1
1
1
1
=
− + − + − + ··· +
−
+
− +
−
2 1 3 2 4 3 5
n−3 n−1 n−2 n n−1 n+1
1 3 1
1
=
− −
2 2 n n+1
n
X
Now, let’s take the limit:
n
∞
X
X 1
1
=
lim
k 2 − 1 n→∞
k2 − 1
k=2
k=2
1 3 1
1
= lim
− −
n→∞ 2 2
n n+1
3
=
4
(c)
2k
k=0 k!
P∞
This seems impossible, but note that
∞ k
X
x
k=0
k!
= ex
So:
∞
X
2k
k=0
k!
= e2
6. Determine if each series diverges, converges conditionally, or converges absolutely.
P
k
(a) ∞
k=1 (−0.25)
This one will converge absolutely, but let’s show that.
∞ ∞ k
X
X
1
(−0.25)k =
4
k=1
That converges (it’s a geometric series), so
(b)
P∞
k k
k=1 (−1) k−1
k=1
P∞
k
k=1 (−0.25)
converges.
Final Exam Review
Math 1220
k
Since limk→∞ k−1
= 1, this series diverges (by the nth term test).
(c)
P∞
1
k=1 k sin k
lim k sin
k→∞
1
1
= lim x sin
k x→∞
x
1
= lim+ sin x
x→0 x
sin x
= lim+
x
x→0
=1
So this series diverges by the nth term test.
7. If you want to use Sn to approximate the sum
the error to be less than 0.001?
Pn
k=1
(−1)k
k ,
what do you need to choose for n in order for
In the language of the Alternating Series Test, an =
1
sum is n+1
. So:
1
k,
so the error in using Sn to approximate the
1
< 0.001
n+1
1
1
<
n + 1 1000
1000 < n + 1
n > 999
8. Determine the convergence set for the power series
Pn
k=1
(x−1)n
n .
Let’s start by using the absolute ratio test:
an+1
|x − 1|n+1 n
= lim
n→∞ an
n→∞ n + 1 |x|n
n
= lim |x − 1|
n→∞
n+1
= |x − 1|
lim
This is less than 1 if x ∈ (0, 2). Now let’s try the endpoints:
n
X
(2 − 1)n
k=1
n
=
=
n
X
1n
k=1
n
X
k=1
n
1
n
Final Exam Review
Math 1220
So it diverges when x = 2.
n
X
(0 − 1)n
k=1
n
=
n
X
(−1)n
k=1
n
This converges by the alternating series test, since limn→∞ n1 = 0. The convergence set is thus [0, 2).
9. Find the terms of the Taylor series for each of the following functions through (x − a)5 . If you wish,
1
you may use the Taylor series representations of sin(x), cos(x), ln(x), ex , and 1−x
in order to do these
problems faster.
(a) f (x) =
1
;
(1−x)2
a=0
Here’s all the terms:
∞
X
1
=
xn
1−x
n=0
∞
d X n
d 1
=
x
dx 1 − x dx
n=0
∞
X
1
d n
=
x
2
dx
(1 − x)
n=1
=
∞
X
nxn−1
n=1
=
∞
X
(n + 1)xn
n=0
(b) f (x) = sin(x) cos(x); a = 0
This one’s also easy to get all the terms for:
1
sin(2x)
2
∞
1 X (−1)n
=
(2x)2n+1
2
(2n + 1)!
sin(x) cos(x) =
n=0
=
∞
1 X (−1)n 22n+1
2
n=0
(2n + 1)!
(c) f (x) = ex ; a = 1
Again, I’ll compute all of the terms. We know that
f (n) (x) = ex
x2n+1
Final Exam Review
Math 1220
So
f (n) (a)
e
=
n!
n!
And so our series is
ex =
∞
X
e
(x − 1)n
n!
n=0
10. Solve the differential equation
dy y
− = 3x3
dx x
if y = 3 when x = 1.
This is a first-order linear differential equation, so it’s of the form
dy
+ P (x)y = Q(x)
dx
with integrating factor
R
e
P (x) dx
R
=e
− 1x dx
= e− ln x
= eln x
−1
= x−1 =
1
x
So let’s do this. Multiply both sides by 1x :
dy y
− = 3x3
dx
x#
"
1 dy y
−
= 3x2
x dx x
#
Z
Z "
1 dy y
−
dx = 3x2 dx
x dx x
1
y = x3 + C
x
y = (x3 + C)x
Now plug in the initial conditions:
3 = (13 + C)1
3 = 1+C
2=C
Final Exam Review
Math 1220
So the solution is:
y = (x3 + 2)x
11. Given the differential equation
dy
dx
= 1 − y 2 , compute:
Here’s the slope field:
2
1
0
-1
-2
0
1
2
3
4
(a) limx→∞ y if y = 0 when x = 0
It looks from the slope field like the limit is 1.
(b) limx→∞ y if y = 10 when x = 0
It looks from the slope field like the limit is 1.
(You may want to plot a slope field.)
12. Use Euler’s method to approximate a solution to the differential equation
y 0 = −y
over the interval [0, 1] with h = 0.2 if y = 1 when x = 0.
n
0
1
2
3
4
5
xn
yn
0
1=1
0.2
1 + 0.2(−1)=0.8
0.4
0.8 + 0.2(−0.8)=0.64
0.6
0.64 + 0.2(−0.64)=0.512
0.8
0.4096 + 0.2(−0.4096)=0.32768
1.0 0.32768 + 0.2(−0.32768)=0.262144
13. (Challenge) Represent 0.099 as a fraction.
Final Exam Review
Math 1220
We can represent 0.099 as
∞
X
∞
X
99
99
99
1 n−1
99
1
99
11
=
=
=
=
=
1
1000n
1000 1000
1000 1 − 1000
1000 − 1 999 111
n=1
n=1
14. (Challenge) Given f (x) = sin(x):
(a) Find the Taylor polynomial of order 5 based at a = 0.
The first five derivatives of f (x) = sin(x), evaluated at 0, are:
f (0) = sin(0) = 0
f 0 (0) = cos(0) = 1
f 00 (0) = − sin(0) = 0
f 000 (0) = − cos(0) = −1
f (4) (0) = sin(0) = 0
f (5) (0) = cos(0) = 1
So the Taylor polynomial is:
1
0
−1 3 0 4 1 5
· x + x2 +
x + x + x
1!
2!
3!
4!
5!
1 3 1 5
≈x− x + x
3!
5!
f (x) ≈ 0 +
(b) Find a formula for R5 (x) (i.e. the remainder).
The remainder formula is
R5 (x) =
f (6) (c) 6 − sin(c) 6
x =
x
6!
6!
(c) Use your formula for R5 (x) to find an upper bound on the value of sin
π
4
Taylor’s Formula tells us that:
sin(π/4) = π/4 −
sin(c)
1
1
(π/4)3 + (π/4)5 −
(π/4)6
3!
5!
6!
Final Exam Review
Math 1220
Where c is between x and a, i.e. c ∈
when c = 0:
h
i
0, π4 . Since sin(x) is increasing, the upper bound will occur
sin(0)
1
1
(π/4)3 + (π/4)5 −
(π/4)6
3!
5!
6!
1
1
0
< π/4 − (π/4)3 + (π/4)5 − (π/4)6
3!
5!
6!
1
1
sin(π/4) < π/4 − (π/4)3 + (π/4)5
3!
5!
< π/4 −