Math 156 Fall 2013
Quiz 3 Solutions
We use integration by parts
Z
udv = uv −
Z
vdu.
to evaluate each of the following.
1:
Z
xe−x dx.
Solution. Let u = x and dv = e−x dx. Then v = −e−x and use integration by parts,
Z
2:
Z
−x
xe dx = −xe
−x
+
Z
e−x dx = −xe−x − e−x + C.
tan−1 (3x)dx.
Solution. Let u = tan−1 (3x) and dv = dx. Then v = x, du =
by parts (use w = 1 + 9x2 , dw = 18xdx in the second step)
Z
3:
Z
3
tan−1 (3x)dx = x tan−1 (3x) −
2
x ln(x )dx = 2
Z
Z
4:
Z
π
4
Z
x3 ln(x)dx =
and use integration
ln(1 + 9x2 )
3xdx
−1
=
x
tan
(3x)
−
+ C.
1 + 9x2
6
x3 ln(x)dx. (Use the property of logarithm ln(x2 ) = 2 ln(x)).
Solution. Let u = 2 ln(x) and dv = x3 dx. Then u =
parts
2
3dx
,
1+(3x)2
x4
,
4
du =
2dx
,
x
and use integration by
x4 ln(x) Z x4 2dx
x4 ln(x) Z x3
x4 ln(x) x4
−
·
=
−
dx =
−
+ C.
2
4
x
2
2
2
8
t sin(2t)dt.
0
Solution. Let u = t and dv = sin(2t)dt. Then v = − cos(2t)
, du = dt, and use integration
2
by parts (use the facts that cos( π2 ) = 0 = sin(0), cos(0) = 1 = sin( π2 ))
Z
0
π
4
π
π
π
−t cos(2t) 4 Z 4 − cos(2t)
sin(2t) 4
1
t sin(2t)dt =
−
dt = 0 +
= .
2
2
4 0
4
0
0
Grade Distribution of this quiz:
Meaning of the scores:
9, 10 = Very good, familiar with the related materials and skillful, with minimal computational errors. Keep on!
8 = good, familiar with most of the related materials, with a few computations errors.
Make an effort to do better.
7 = Passing, less familiar with the related materials and more computational errors and
algebraic errors. We have lots of room to improve.
6 = Borderline. We need to catch it up. If you have trouble doing your homework, it
might be time to visit your instructor to get help. Do not wait to let the trouble accumulate.
At most 5 = This might be a dangerous warning signal. We are failing! It should definitely be the time for us to see the instructor and get assistance to understand the materials
and to practice MORE.
Scores
Frequency
Percentage
9 < and ≤ 10 8 < and ≤ 9
7
2
18.9
5.4
7 < and ≤ 8 6 < and ≤ 7 5 < and ≤ 6 ≤ 5
8
7
4
9
21.6
18.9
10.8
24.4
Discussions and Comments
(1) Problem: Differentiation and integration error are still the number one trouble for us.
The followings are the most frequent in this quiz.
Errors
u = ln(x2 ) and du = dx
x2
u = tan−1 (3x) and du = 1/(1
√ + 3x)dx
−1
u = tan (3x) and du = x/ x2 − 1dx
dv
= t and v = 1
R
tan−1 xdx = 1/(1 + x2 )
R
tan−1 xdx = 1/(1 + x2 )
Correct Way
du = 2xdx
x2
3dx
du = 1+9x
2
3dx
du = 1+9x
2
2
v = t2
use by parts with u = tan−1 (x) and dv = 1
use by parts with u = tan−1 (x) and dv = 1
What to do: Once again, I urge everyone to be familiar with the chain rule, differentiation
formulas. Make up your own table for differentiation rules and remember it.
(2) Problem: Many algebra errors. Some typical ones are listed below.
Errors
tan−1 (x) = tan1 x
1
1
2 = 1 + 9x2
R1+9x−x
R
R −x
xe
dx
=
xdx
e Rdx
R 3
R 3
2
x ln(x )dx = x dx ln(x2 )dx
Correct Way
tan−1 (x) is the inverse function of tan(x)
use substitution w = 3x
use by parts with u = x
use by parts with u = ln(x2 )
What to do: For the integration errors, review Properties of Integrals in Section 5.2, Page
270 of text.
(3) Problem: Several of us did not get their solutions right mainly because they did not
know what is the correct version of integration by parts. They wrote the by-parts formula
on the solution sheet but their formula is wrong!
What to do: If we are to use integration by parts, we may write down
Z
and follow it to do the problems.
udv = uv −
Z
vdu