Math 116 CALCULUS II
REVIEW OF LECTURES – X
June 14 (Tue), 2016
Instructor:
Yasuyuki Kachi
Line #: 81300.
• Integration by parts.
By now you must be a little tired of hearing: Ninety nine percent of indefinite
integrals is “not feasible”. There are one percent of indefinite integrals that are
“feasible”. You already know it. You also know what precisely the word “feasible”
means, as I made clear of it. Speaking of “feasible”, a while ago, I told you:
“There are two common rules
‘feasible’ 1%”.
or, methods we can utilize to do those
“Discovery of those two methods ‘made’
ball-park
integrals feasible.”
1% of indefinite
I told you the names of those two methods:
“Substitution integral method”,
and
“Integration by parts method”.
I took care of the former. Today I want to do the latter. We are now concerned with
indefinite integrals. So, we move back to indefinite integrals.
• So, “Integration by Parts”. Let us have a glance at the following:
Z
x e2x dx.
Or, if you like it better, the same question can be phrased as
“Find an antiderivative of
1
f x
= x e2x .”
Do you think this is “feasible”? If so, do you know the answer right way? For this, I
actually know the answer. My answer is, “yes, it is indeed feasible”. I can concretely
write it out as
Z
1
1 2x
e
+ C.
x −
(∗)
x e2x dx =
2
4
You say “how do you know?” Well, actually, I did some work earlier. But from
your point of view, the substitution integral method does not seem to work for this
particular integral. So, how did I really get this answer? Well, I don’t want to tell
you. I would rather keep it as a “secret”.
Oh, please don’t be upset. Because you can be the judge. I might be lying to
you. I might have just randomly made it up and trying to sell that this is the correct
answer, but it might really be a wrong answer: You have every reason to be suspicious
about what I say, because I am refusing to show you how to get the answer, right?
Right. So, I let you be the judge. Please decide whether I am lying. But the question
is how you can decide whether I am lying, because you are not told of how I have
actually pulled the answer? Yes. You indeed know exactly how. Namely, all you
have to do is “differentiate” my answer:
#
"
1
1 2x
d
e
x −
dx
2
4
"
#
"
#
1
d 1
1 1 d
2x
2x
=
·e
+
·
x−
x−
e
dx
2
4
2
4
dx
Leibniz rule
1
1
1
=
· 2 e2x
· e2x +
x −
2
2
4
=
1 2x
e
+
2
=
1 2x
e
+ x e2x −
2
1
x · 2 e2x −
2
1
4
· 2 e2x
1 2x
e
2
= x e2x .
All right, this outcome matched what was inside the integral symbol in (∗) above.
So, this means you have just confirmed that the above answer of mine was correct.
So, your ruling: I was not lying. Terrific job of officiating.
2
• But this proof doesn’t quite tell you how to pull my answer (∗). This proof was
not a heuristic argument . Aren’t you curious as to know how I pulled my answer?
All right. Let’s now completely forget about what you saw above. Let’s rewind the
tape back. Both you and I pretend that we don’t know the answer to the above
integral (∗). Let’s focus on “how to solve” the following problem:
Question. Find the following indefinite integral
Z
• Key.
x e2x dx.
There is one perfect formula that you can use to an integral like this.
• Integration by parts.
This is the main theme of today’s lecture, all of a sudden. Please memorize :
Formula 1 ( Integration by Parts formula ).
Z
u dv
=
uv −
Z
v du
.
But you don’t understand the meaning of it. No, not yet. Yet as I say, please
memorize it regardless, no matter what, and whatever it takes. The right approach
for you in this case is, first memorize it, then next listen to my deciphering of it.
• The above is actually a formula. But it does not look like one. Your first impression
might be “What the heck is that?” What is u, what is v? Actually, u and v are
both some quantities involving another letter x . Thus, you can use, for example,
ϕ and ψ istead, or even ♥ and ♦ instead. There is no logical reason to use the
letters u and v. Somehow u and v are a popular choice for the pair of letters that
is used to spell out this formula. We have no compelling reason to go against it.
3
Z
Okay, fine. But how does the formula like
=
u dv
uv −
Z
v du
help us evaluate something like
Z
(∗)
x e2x dx ?
Yes, I will tell you that. It takes four small steps Steps 0, 1, 2 and 3 below .
Step 0 (Preliminary).
Set u and dv as follows:
Z
x
e2x dx
x y x y
u
dv
thus
x = u
d v = e2x dx
u=x
Create the chart table :
Step 1.
,
.
dv = e2x dx
We need to complete the chart. More precisely, we need to express
du
and
v
each, in terms of x and dx . Don’t ask me why. Just do as I say and see what
happens. First, figuring out d u is super-easy, because u = x just implies
du = dx.
Second, figuring out v is not too hard, indeed, based on d v = e2x , this
question is essentially just the question of finding an antiderivative of e2x . We
1 2x
certainly know that one antiderivative of e2x is
e .
2
4
So, voilà:
u=x
du = dx
Oh, that wasn’t too bad.
Step 2.
v =
1
2
e2x
dv = e2x dx
Now this is the ‘fun part’. Write out “Integration by Parts Formula”
Z
=
u dv
uv −
Z
v du
from page 3 . In it, substitute the data in the chart in Step 1:
(#)
Z
2x
x
e
x y x
y
u
Here,
=
dx
dv
1
e2x
x
2
x y x y
u
v
−
Z
1
dx.
e2x
2
x
y x y
v
du
all I did was to substitute all the data in the chart into the formula
in the box .
All right. What have we got here? Let me wipe off all of u, v, du and dv
underneath the line (#) and duplicate the line:
(#)
Z
2x
xe
dx
=
Z 1
1 2x 2x
x
e
−
e
dx.
2
2
The left-hand side integral in (#) is what we want to evaluate. It now equals the
right-hand side of (#). The right-hand side of (#) has got two terms, one of which
still involves an integral. So you might say
5
“What are we really doing here? We were in the process of evaluating one
integral. We were told to follow your steps and along the way another
integral showed up. Are we then supposed to deal with that new integral,
the one that just showed up? But this looks like a ‘loop’, or ‘Catch 22’.
Are you saying in order to evaluate one integral you need to evaluate
another integral? This sounds a little absurd to me.”
Okay. You made a strong argument. But I say this:
“Yes, you are absolutely right, when evaluating one integral, we sometimes
need to deal with another integral, as a necessary step. But I have
good news for you: This is not a ‘loop’ or ‘Catch 22’. Indeed, take a
look at the new integral more closely:
Z 1 2x e
dx.
2
This integral is something we know how to handle. This one is doable.”
Yes indeed. We all know how to do this last integral. Let’s do it together. And
that’s ‘Step 3’:
Z
Z 1
1 2x e2x dx
dx =
Step 3.
e
2
2
1 2x
1
·
e
+ C
=
2
2
1 2x
e
+
4
=
1
C .
2
1
1
C as C. In fact, technically you should call
C as C ′ .
Now, rewrite
2
2
However, once you have done that, there is no C any more so you can just re-name
C ′ as C. So in short,
(&)
Z 1 2x dx
e
2
=
6
1 2x
e
+ C.
4
Z
Now, this is not the final answer for
x e2x dx.
Indeed, this is a portion
of the answer, because, once again,
(#)
Z
2x
xe
Z 1
1 2x 2x
x
−
dx .
e
e
2
2
x
y
=
dx
(&)
We just did the (&) part. The entire right-hand side of (#) is what we want. We
1 2x
found out the (&) part to be
e
+ C. So then
4
1
1 2x
2x
−
e
e
+ C
x
2
4
would be the answer. Yes indeed. We can delete the parenthesis and write it as
x
1
e2x
−
2
1 2x
e
− C.
4
Here, − C appeared at the tail-end, but youcan just replace it with + C. In fact,
even though similarly to the previous time technically you should call − C as
C ′ , once you have done that, there is no longer C so you can just re-name C ′ as
C. So
1
2x
−
e
x
2
1 2x
e
+ C
4
is your final answer. Let’s highlight this final answer:
Z
x e2x dx
=
x
1
−
e2x
2
1 2x
e
+ C.
4
This outcome is exactly the same as the answer I gave you earlier. Right?
7
Whew. This solution was long. But I broke down the steps and that’s why it
appeared long. We can combine all the steps, and can write it down in a relatively
short form see ‘Solution a’ in the next page , but before learning how to do that,
we should first look back what we did.
• In retrospect..
Let’s look back what we did one more time. We tried to solve
Z
x e2x dx
the original integral .
But we initially had no clue.
But then somehow we relied on a formula, called “Integration by Parts Formula”,
and that rescued us. Namely, it enabled us to convert this original integral into
something like
Z 1
1 2x 2x
e
−
e
dx.
x
2
2
So, by applying the “Integration by Parts Formula”, we have converted one integral
“Z ”
into another, well, not quite, but into a sum of a term without
and a term
“Z ”
“Z ”
with
. Most importantly, we knew how to handle the term with
namely, the integral that showed up in that term was the kind we have already
practiced. That way we were able to reach our goal, namely, to evaluate the original
integral.
So, the morale of the story is, by relying on the “Integration by Parts Formula”
we were able to convert the original integral, which was difficult, into another, easier,
integral.
8
• Below is a ‘veteran’ way to write the whole solution :
Solution a.
Z
2x
x
e
x y x
u
dx
y
dv
1
=
x
e2x
2
x y x y
u
−
v



1
e2x
dx
2
x
y x y
v
u=x
du = dx
v =
1
2
e2x
dv = e2x dx
=
1
x e2x −
2
1
2
=
1
x e2x −
2
1 2x
e
+ C
4
1
x −
2
• Let’s do more examples.
Let us find
Z
ln x dx.
Here is how it goes.
9
du



1 2x
e dx
2
=
1
x e2x −
2
=
Example 1.
Z
Z
·
1 2x
e
+ C
2
1 2x
e
+ C.
4
Step 0 (Preliminary).
Set u and dv as follows:
Z
ln x
dx .
x y x y
u
dv
thus
u = ln x
Step 1.
,
d v = dx
.
Rewrite this integral using Formula 1 in page 3.
We need to
know what d u is, and what v is .
For that matter, create the chart
u = ln x
Step 2.
Complete the chart:
dv = dx
dx u = ln x
du =
Step 3.
(#)
1
x
Using these,
Z
ln x
dx
x y x y
u
dv
=
ln x
x
u
v =x
dv = dx
· x
y x y
v
10
−
Z
x ·
x y
v
1
x
dx .
x du
y
Step 4 (Finishing Touch).
(#)
Z
ln x
dx
Continue from (#):
=
ln x
=
=
x ln x
ln x
· x
Z
−
·x
−
Z
x ·
1
x
dx
1 · dx
x + C.
−
• Below is a ‘veteran’s’ way :
Z
ln x
dx
x y x y
u
=
ln x · x
x y x y
dv
u
−
v



Z
du =
=
x ln x − x + C.
1 dx
• Let’s do another example:
Example 2. Find
11
1
x2
dx u = ln x
x ln x −
ln x ·
1
dx
x ·
x
x y x y
v
=
Z
Z
dx.
1
x
du
v =x
dv = dx



Step 0 (Preliminary).
Set u and dv as follows:
Z
ln x
1
x2
·
x y
dx .
x
y
u
dv
thus
u = ln x
Step 1.
,
1
dx
x2
dv =
.
Rewrite this integral using Formula 1 in page 3.
We need to
know what d u is, and what v is .
For that matter, create the chart
u = ln x
Step 2.
dv =
du =
(#)
Z
dx
Complete the chart:
dx u = ln x
Step 3.
1
x2
1
x
v =−
dv =
1
x2
1
x
dx
Using these,
1
ln x ·
dx
=
ln
x
· −
x2
x y x y
x y x
u
dv
u
12
1
x
!
y
v
−
Z
1
−
x
x v
!
1
dx.
x
y x y
·
du
Step 4 (Finishing Touch).
(#)
Z
ln x ·
Continue from (#):
1
dx =
ln x ·
x2
1
−
x
−
ln x
x
=
−
ln x
x
=
−
ln x + 1
x
=
+
+
!
−
1
x2
Z
−
1
−
x
Z
!
·
1
dx.
x
dx
1 + C
x
+ C.
• Below is a ‘veteran’s’ way :
Z
!
!
Z
1
1
1
1
−
·
−
dx
=
ln
x
·
−
dx
ln x ·
x2
x
x
x
x y x y
x y x y
x y x y
u
dv
u
v
v





=
−
ln x
x
=
−
ln x
x
=
−
ln x + 1
x
+
+
13
u = ln x
du =
1
dx
x
1
x2
Z
−
v =
dv =
dx
1 + C
x
+ C.
du
−1
x
1
dx
x2





• How about the next one:
Example 3. Find
Z Step 0 (Preliminary).
ln x
2
dx.
Set u and dv as follows:
Z
2
ln x
dx .
x y x y
u
dv
thus
u =
Step 1.
ln x
2
,
d v = dx
Rewrite this integral using Formula 1 in page 3.
know what d u is, and what v is .
For that matter, create the chart
u=
Step 2.
.
2
ln x
Complete the chart:
u=
du =
2
ln x
2 ln x
x
dx 14
dv = dx
v= x
dv=dx
We need to
Step 3.
Using these,
Z
(#)
ln x
x
2
u
dx
=
y x y
x dv
(#)
ln x
2
−
2 ln x
x
x
x ·
x y
v
v
dx .
y
du
Continue from (#):
=
dx
·x
Z
y x y
u
Step 4 (Finishing Touch).
Z
ln x
2
=
ln x
ln x
2
2
·x
·x
Z
−
Z
−
x ·
2
ln x
2 ln x
x
dx
dx.
Now, we need to finish this off. For that matter, we need to know how to deal with
Z
2 ln x dx.
Guess what? This is essentially what we have already dealt with earlier. Namely, in
Example 1, we had
Z
ln x dx = x ln x − x + C.
So
Z
ln x
2
dx
=
=
=
ln x
x
x
2
ln x
ln x
·x −
2
2
− 2
Z
− 2x
15
2
ln x
dx
x ln x − x
ln x
+ C
+ 2 x + C.
• Below is a ‘veteran’s’ way :
Z
2
ln x
dx
x y x y
u
=
dv
2
ln x
·x
x y x y
u
−
v

u=





du =
2
ln x
=
x
ln x
=
x
ln x
=
Z
x ·
x y x
v
2
ln x
2 ln x
x
Z
·x −
dx 2
2
− 2
2
− 2x
ln x
e−x
x
+
e−x
x2
ln x
dx.
Split the given integral as
Z
e−x
dx
x
+
16
Z





dx
e−x
dx
x2
+ C
+ 2 x + C.
This is a little tricky. Somehow, the following method works.
Method.
x ln x − x
Find
Z 
dv=dx
dx
y
du
v= x
• Next, why don’t we try something with a different flavor.
Example 4.
2 ln x
x
.
Try to evaluate the first integral
e−x
dx,
x
Z
by applying the integration by parts formula. Let’s see how it goes. Below is ‘sameold, same-old’.
Step 0 (Preliminary).
First rewrite this last intergral as
Z
1
x
· e−x dx.
Set u and dv as follows:
Z
1
· e−x dx .
x
x y x y
u
dv
thus
u =
Step 1.
1
x
d v = e−x dx
,
.
Rewrite this integral using Formula 1 in page 3.
know what d u is, and what v is .
For that matter, create the chart
u=
1
x
17
dv = e−x dx
We need to
Step 2.
Complete the chart:
u=
du =
Step 3.
(#)
Z
1
x
−1
dx
x2
Using these,
v = − e−x
dv = e−x dx
Z 1
−1
1
−x
−x
−
− e−x ·
· e dx =
·
−e
dx .
x
x
x2
x y x y
x y x y
x y x y
u
dv
u
Step 4 (Finishing Touch).
Z
1
1
· e−x dx =
x
x
=
v
v
du
Continue from (#):
Z −x
−x
·
−
−e
·
−e
−
e−x
x
−
−1
x2
dx
e−x
dx.
x2
Z
All right. Let’s take a break here, and see what we have got. In one line, it is
Z
1
x
−x
· e
=
dx
−
e−x
x
−
Z
e−x
dx
x2
.
What have we got here? As a result of applying the integration by parts formula to
the integral in the left-hand side of the above box, we arrived that the right-hand
side of the box. Now, we don’t seem to know how to handle the new integral in the
right-hand side of the box. But hey, here is some good news. Why don’t we shift
that integral in the boxed identity to the left-hand side? Then, all of a sudden:
Z
e−x
dx +
x
Z
e−x
dx
x2
18
=
−
e−x
x
+ C.
The left-hand side of this new identity can be written as
Z e−x
x
e−x
x2
+
dx,
and you know what, this is exactly what we wanted to evaluate. So we are done:
Z e−x
x
+
e−x
x2
dx
19
=
e−x
−
x
+ C.