Math 116 CALCULUS II
REVIEW OF LECTURES – V
June 7 (Tue), 2016
Instructor:
Yasuyuki Kachi
Line #: 81300.
• Antiderivatives.
Today we embark on exploring the new subject antiderivatives . The spelling:
antiderivative,
anti-derivative.
or
As an operation, antiderivative is just “the inverse” of the derivative operation. So,
conceptually this is not something that is entirely new. However, this does not mean
that the level of difficulty of the two subjects — the derivatives and the antiderivaties
— is the same. Indeed, you will probably find the subject of antiderivatives as harder,
by a few notches. Please see Remark. The first thing I will do as usual : I will
cover the definition. As far as the definition goes, it is quite simple, and easy. You see
the definition below and might be misled that antiderivatives are easy. The reality
is, some are easier than others. Today we will only see those of the easiest kind. We
all know that 2 x is the derivative of x2 . Then we say x2 is an antiderivative
of 2 x. More generally:
Definition
If
(Antiderivatives).
1f x
call F x an antiderivative of f x .
is the derivative of F x , then we
Oh, is that it? Yes indeed, that is it. But you will have to see lots of basic examples
also, to get a sense:
Example
F x
F x
F x
F x
F x
1 Here
1.
= x2 is an antiderivative of f x
= x3 is an antiderivative of f x
= x4 is an antiderivative of f x
= x5 is an antiderivative of f x
= x6 is an antiderivative of f x
= 2 x.
= 4 x3 .
= 6 x5 .
= 3 x2 .
= 5 x4 .
I did not say “the” antiderivative. As for the reason, see page 10–page 11.
1
• By the way,
F x
= x is an antiderivative of f x
= 1.
Example 2.
F x
=
1 2
x is an antiderivative of f x = x.
2
F x
=
1 3
x is an antiderivative of f x = x2 .
3
F x
=
1 4
x is an antiderivative of f x = x3 .
4
F x
=
1 5
x is an antiderivative of f x = x4 .
5
F x
=
1 6
x is an antiderivative of f x = x5 .
6
Example 3.
F x
= x−1 is an antiderivative of f x
= − x−2 .
F x
= x−2 is an antiderivative of f x
= −2 x−3 .
F x
= x−3 is an antiderivative of f x
= −3 x−4 .
F x
= x−4 is an antiderivative of f x
= −4 x−5 .
F x
= x−5 is an antiderivative of f x
= −5 x−6 .
2
Paraphrase of Example 3:
Example 4.
F x
=
1
x
F x
=
1
x2
is an antiderivative of f x
=
−2
.
x3
F x
=
1
x3
is an antiderivative of f x
=
−3
.
x4
F x
=
1
x4
is an antiderivative of f x
=
−4
.
x5
F x
=
1
x5
is an antiderivative of f x
=
−5
.
x6
is an antiderivative of f x
=
1
.
x2
is an antiderivative of f x
−1
.
x2
=
Example 5.
F x
=
−1
x
F x
=
−1
2 x2
is an antiderivative of f x
=
1
.
x3
F x
=
−1
3 x3
is an antiderivative of f x
=
1
.
x4
F x
=
−1
4 x4
is an antiderivative of f x
=
1
.
x5
F x
=
−1
5 x5
is an antiderivative of f x
=
1
.
x6
3
Example 6.
F x
= x1/2 is an antiderivative of f x
=
1 −1/2
x
.
2
F x
= x3/2 is an antiderivative of f x
=
3 1/2
x .
2
F x
= x5/2 is an antiderivative of f x
=
5 3/2
x .
2
F x
= x7/2 is an antiderivative of f x
=
7 5/2
x .
2
F x
= x9/2 is an antiderivative of f x
=
9 7/2
x .
2
Example 7.
F x =
2 x1/2
is an antiderivative of f x
= x−1/2 .
F x
=
2 3/2
x
is an antiderivative of f x = x1/2 .
3
F x
=
2 5/2
x
is an antiderivative of f x = x3/2 .
5
F x
=
2 7/2
x
is an antiderivative of f x = x5/2 .
7
F x
=
2 9/2
x
is an antiderivative of f x = x7/2 .
9
Example 8.
F x
= x−1/2 is an antiderivative of f x
=
−1 −3/2
x
.
2
F x
= x−3/2 is an antiderivative of f x
=
−3 −5/2
x
.
2
F x
= x−5/2 is an antiderivative of f x
=
−5 −7/2
x
.
2
F x
= x−7/2 is an antiderivative of f x
=
−7 −9/2
x
.
2
4
Example 9.
F x
= − 2 x−1/2 is an antiderivative of f x
F x
=
−2 −3/2
x
is an antiderivative of f x = x−5/2 .
3
F x
=
−2 −5/2
x
is an antiderivative of f x = x−7/2 .
5
F x
=
−2 −7/2
x
is an antiderivative of f x = x−9/2 .
7
= x−3/2 .
• Paraphrase (one each from Examples 6, 7, 8, 9).
◦
Paraphrase of the first line in Example 6:
F x
◦
√
x is an antiderivative of f x
=
2
1
√
x
.
= 2
√
x is an antiderivative of f x
1
.
= √
x
Paraphrase of the first line in Example 8:
F x
◦
=
Paraphrase of the first line in Example 7:
F x
◦
1
= √
x
is an antiderivative of f x
=
−1
√
.
2x x
Paraphrase of the first line in Example 9:
F x
=
−2
√
x
is an antiderivative of f x
5
=
x
1
√
x
.
Example 10.
F x
= ex is an antiderivative of f x
F x
= e2x is an antiderivative of f x
= 2 e2x .
F x
= e3x is an antiderivative of f x
= 3 e3x .
F x
= e4x is an antiderivative of f x
= 4 e4x .
F x
= e5x is an antiderivative of f x
= 5 e5x .
= ex .
Example 11.
F x
=
1 2x
e
is an antiderivative of f x = e2x .
2
F x
=
1 3x
e
is an antiderivative of f x = e3x .
3
F x
=
1 4x
e
is an antiderivative of f x = e4x .
4
F x
=
1 5x
e
is an antiderivative of f x = e5x .
5
6
Example 12.
F x
= e−x is an antiderivative of f x
F x
= e−2x is an antiderivative of f x
= −2 e−2x .
F x
= e−3x is an antiderivative of f x
= −3 e−3x .
F x
= e−4x is an antiderivative of f x
= −4 e−4x .
F x
= e−5x is an antiderivative of f x
= −5 e−5x .
= − e−x .
Example 13.
F x
= − e−x is an antiderivative of f x
F x
=
−1 −2x
e
is an antiderivative of f x = e−2x .
2
F x
=
−1 −3x
e
is an antiderivative of f x = e−3x .
3
F x
=
−1 −4x
e
is an antiderivative of f x = e−4x .
4
F x
=
−1 −5x
e
is an antiderivative of f x = e−5x .
5
• More non-trivial ones:
7
= e−x .
Example 14.
2
F x
= ex
F x
= ex
F x
= ex
F x
= ex
3
4
5
2
is an antiderivative of f x
= 2 x ex .
is an antiderivative of f x
= 3 x2 ex .
is an antiderivative of f x
= 4 x3 ex .
is an antiderivative of f x
= 5 x4 ex .
3
4
5
Example 15.
2
F x
= e−x
F x
= e−x
F x
= e−x
F x
= e−x
3
4
5
2
is an antiderivative of f x
= −2 x e−x .
is an antiderivative of f x
= −3 x2 e−x .
is an antiderivative of f x
= −4 x3 e−x .
is an antiderivative of f x
= −5 x4 e−x .
3
4
5
Example 16.
F x
=
1 x2
2
e
is an antiderivative of f x = x ex .
2
F x
=
1 x3
3
e
is an antiderivative of f x = x2 ex .
3
F x
=
1 x4
4
e
is an antiderivative of f x = x3 ex .
4
F x
=
1 x5
5
e
is an antiderivative of f x = x4 ex .
5
8
Example 17.
F x
=
−1 −x2
2
e
is an antiderivative of f x = x e−x .
2
F x
=
−1 −x3
3
e
is an antiderivative of f x = x2 e−x .
3
F x
=
−1 −x4
4
e
is an antiderivative of f x = x3 e−x .
4
F x
=
−1 −x5
5
e
is an antiderivative of f x = x4 e−x .
5
Example 18.
F x
= ln x is an antiderivative of f x
1
.
x
=
• Can I quiz you?
Exercise.
(1)
True or false :
F x
(2)
2x
is an antiderivative of f x
=
1
.
x
= ln
3x
is an antiderivative of f x
=
1
.
x
is an antiderivative of f x
=
1
.
x
True or false :
F x
h
= ln
True or false :
F x
(3)
= ln
4x
Answers for (1), (2), (3)
i
:
All True.
9
Can you explain why?
Example 19.
F x
F x
F x
F x
=
=
=
=
Exercise.
ln x
2
ln x
3
ln x
4
ln x
5
is an antiderivative of f x
is an antiderivative of f x
is an antiderivative of f x
is an antiderivative of f x
=
=
=
=
2
3
4
5
ln x
x
ln x
x
ln x
x
ln x
x
.
2
.
3
.
4
.
Explain why each of the four lines in Example 19 is true, using
d
d α
α−1
= α♥
♥
♥
dx
dx
α is a constant
.
Example 20.
F x
F x
F x
F x
=
1
2
=
1
3
=
1
4
=
1
5
ln x
2
ln x
3
ln x
4
ln x
5
is an antiderivative of f x
is an antiderivative of f x
is an antiderivative of f x
is an antiderivative of f x
10
=
.
2
.
3
.
4
.
ln x
x
ln x
=
ln x
=
ln x
=
x
x
x
Example 21.
=
is an antiderivative of f x
= 4x
is an antiderivative of f x
= 5x
is an antiderivative of f x
F x
= 2x
F x
= 3
F x
F x
ln 2
2x .
=
ln 3
3x .
=
ln 4
4x .
=
ln 5
5x .
is an antiderivative of f x
x
Example 22.
F x
=
1
ln 2
2x
is an antiderivative of f x
= 2x .
F x
=
1
ln 3
3x
is an antiderivative of f x
= 3x .
F x
=
1
ln 4
4x
is an antiderivative of f x
= 4x .
F x
=
1
ln 5
5x
is an antiderivative of f x
= 5x .
Exercise.
22 is true.
Based on Example 21, explain why each of the four lines in Example
Example 23.
F x
= ln
ln x
is an antiderivative of f x
11
=
1
.
x ln x
• Antiderivative of a quantity that involves more than one term.
Let’s ‘mix’ what we have already learned. Namely:
Example 24.
1 3
x + x
3
2 5
x − 2 x3 + 4 x2
5
√
√
2
x x + 2 x
3
ex − e−x
1 2x
e +
2
1
x
ln
1 3x
e
3
ln x
+
ln x
√
is an antiderivative of
is an antiderivative of
2
1
.
x + √
x
ex + e−x .
is an antiderivative of
2 x4 − 6 x2 + 8 x.
is an antiderivative of
is an antiderivative of
+ ln x
x2 + 1.
is an antiderivative of
−
1
x2
e2x + e3x .
+
is an antiderivative of
1
.
x
1
x ln x
2
+
• Antiderivative is not unique. Two antiderivatives of f x
constant.
ln x
“ If F x is an antiderivative of f x , then
F x − 3,
F x +
1
,
2
too are all antiderivatives of the same f x . ”
12
√
F x − 2,
.
differ by a
Here is one important thing you should know.
F x + 1,
x
etc.
• If you are not sure of what I mean by that, take a look at the following:
Example 25.
F x
=
In the above we have seen:
1 3
x is an antiderivative of f x = x2 .
3
This is true. But then the following is also true:
F x
=
1 3
x + 1 is an antiderivative of f x = x2 .
3
More generally, the following is true:
F x
=
1 3
x + C is an antiderivative of f x = x2 ,
3
as long as C denotes a constant
real number .
1 3
x plus
Now, is there any antiderivative of x2 , other than those, namely, “
3
1 3
a constant”? The answer is no :
x + C where C is a constant that varies
3
exhaust all the antiderivatives of x2 . More generally:
Nature (of the notion of antiderivatives).
“ If F x is one antiderivative of f x , then
F x +C
C is an arbitrary constant
real number
represents all the antiderivatives of f x .”2
2 Actually,
mathematicians are a bit meticulous
about this statement. The underlying premise
which I do not mention in the main text is “f x is defined over an interval ”.
13
• Indefinite Integrals.
All of a sudden, we turn our attention to another the second major topic: The
so-called “ indefinite integrals ”. What are they? Indefinite integrals are basically
just antiderivatives. So, good news: The
rest of today’s lecture is still closely related
to what we have covered so far today. Here, I said “basically”. There is actually a
little more to say than that:
1.
The notion of indefinite integrals is always associated with the “symbol”
Z
.
You have seen this symbol for the first time. Pronunciation:
Z
:
‘‘ integral ’’ .
There is some proper way to signify an antiderivative of f x using this symbol
as
3
to how, I tell you shortly , and you call it the indefinite integral of f x .
2.
The indefinite integral of f x , properly signified using that symbol, must
represent all the antiderivatives of f x . Thus, the answer for the indefinite
integral of f x must come with “+ C”.
The above 1. and 2. explain what indefinite integrals are, relative to antiderivatives.
Well, the above items 1. and 2. sound “technical”.
I purposely included
those
technical items before giving the official definition of indefinite integrals. Putting
those technicalities aside, though, basically speaking, the difference between indefinite
integrals and antiderivatives is, they are the same concept, except there is a difference
in style, and which one you resort to in various mathematical contexts is totally your
call, namely, it is a matter of whether you want to express it using mathematical
symbols or want to express it using words. We used to say in words
“ something is an antiderivative of something. ”
The symbol
Z
allows you to write the same content only using mathematical
symbols. Now I should tell you how exactly to do that.
3 Note:
I said “the indefinite integral”, as opposed to “an indefinite integral”.
14
Definition–Notation (Indefinite Integrals).
Suppose F x is one antiderivative of f x . Then you write
Z
f x dx
=
F x
+ C
.
• We officially call the left-hand side in the above highlighted box the
indefinite integral of f x , with respect to x. We call C the integral constant .
Example.
Now we are free to write something like
Z
x dx
1 2
x + C,
2
=
to mean
“ F x
=
1 2
x is an antiderivative of f x = x. ”
2
Also, we are free to write something like
Z
x2 dx
1 3
x + C,
3
=
to mean
“ F x
=
1 3
x is an antiderivative of f x = x2 . ”
3
• So, we no longer have to say verbally “something is an antiderivative of something”,
even though we can, and often will, whenever we feel like it. What we have just done
is we have created an alternative, shorter, way of expressing it, and it only uses
mathematical symbols, which is to our advantage.
15
• Two remarks are in order, concerning the expression
Z
f x dx
=
F x
+ C
.
1. Always include “+ C” as a part of your answer.
If you look at the right -hand side of the above highlighted box, you see there is
“+ C”. This is necessary , due to the reason I already explained. Many students
who saw the indefinite integrals for the first time tend to omit “+ C” as a part of
the answer. So, please do not forget to include + C.
2. Always include the tail-end “dx” inside the integral symbol.
If you look at the left -hand side of the above highlighted box a little more carefully,
you notice there is this tail-end “dx”. This tail-end dx is necessary , in fact, it is
absolutely indispensable. Again, many students who saw the indefinite integrals for
the first time tend to omit the tail-end dx. Please never do that. Every single time
you omit the tail-end dx, I will correct you. I will be consciously very square about
this when I grade your papers. The reason why this tail-end is absolutely necessary
may not be clear to you at this stage. Actually, at this stage you do not have to
know why. The reason for the necessity to attach “dx” will become
transparent,
once we get to the topic of “substitution integrals” the next lecture . At this stage,
please just know that, attaching the tail-end “dx” is just a convention, and you are
required to follow the convention .
• Let’s familiarize ourselves with this new way of writing, called “the indefinite
integral”.
Exercise 1.
“F x
h
Answer
Use the integral symbol to paraphrase
= x4 is an antiderivative of f x
i
:
Z
4 x3 dx
=
16
= 4 x3 . ”
x4 + C.
Exercise 2.
“F x
h
Answer
Use the integral symbol to paraphrase
=
i
:
h
Answer
h
Answer
= 2
i
:
h
Answer
√
=
−1
x
=
1
. ”
x2
+ C.
x is an antiderivative of f x
Z
√
1
x
=
dx
2
√
= √
1
. ”
x
x + C.
Use the integral symbol to paraphrase
=
i
:
√
2 3/2
x
is an antiderivative of f x =
x. ”
3
Z
Exercise 5.
“F x
1
dx
x2
Use the integral symbol to paraphrase
Exercise 4.
“F x
is an antiderivative of f x
Z
Exercise 3.
“F x
−1
x
√
x dx
2 3/2
x
+ C.
3
=
Use the integral symbol to paraphrase
= ex is an antiderivative of f x
i
:
Z
ex dx
=
17
= ex . ”
ex + C.
Exercise 6.
“F x
h
Answer
Use the integral symbol to paraphrase
=
i
:
Z
Exercise 7.
“F x
h
Answer
h
Answer
e2x dx
1 2x
e
+ C.
2
=
Use the integral symbol to paraphrase
=
i
:
−1 −x2
2
e
is an antiderivative of f x = x e−x . ”
2
Z
Exercise 8.
“F x
1 2x
e
is an antiderivative of f x = e2x . ”
2
2
x e−x dx
−1 −x2
e
+ C.
2
=
Use the integral symbol to paraphrase
= ln x is an antiderivative of f x
i
:
Z
1
dx
x
=
ln x + C.
Exercise 9.
Use the integral symbol to paraphrase
= 2 x4 + 8 x is an antiderivative of f x
i
:
“F x
h
Answer
Z 8 x3 + 8 dx
=
18
1
. ”
x
=
= 8 x3 + 8. ”
2 x4 + 8 x + C.
Exercise 10.
“F x
h
Answer
Use the integral symbol to paraphrase
= x +
i
:
Z Exercise 11.
“F x
=
1
x
is an antiderivative of f x
1 dx
x2
1 −
i
Answer
1 x
e +
2
Exercise 12.
h
Answer
1
x
1
. ”
x2
+ C.
1 −x
e
2
=
is an antiderivative of
1 x
e −
2
1 −x
e . ”
2
:
Z 1 x
e −
2
“F x
x +
= 1 −
Use the integral symbol to paraphrase
f x
h
=
=
i
:
1 −x dx
e
2
=
1 x
e +
2
1 −x
e
+ C.
2
Use the integral symbol to paraphrase
2
1 ln x
2
Z
is an antiderivative of f x
ln x
dx
x
=
2
1 ln x
+ C.
2
=
19
ln x
. ”
x
Exercise 13.
“ F x
h
Answer
Note.
later.
i
Use the integral symbol to paraphrase
= x ln x − x
is an antiderivative of
f x
= ln x. ”
:
Z
ln x dx
=
x ln x − x + C.
You have not seen the content of Exercise 13 before. We will revisit this
• Why is the adjective ‘indefinite’ ? Is there a ‘definite integral’ ?
By the way, you might naturally wonder
“ Why do we have to attach the adjective ‘indefinite’ to the word
‘integral’ ? Is there another kind of an integral, perhaps something
to be called ‘definite integral’ ?”
Yes, you are absolutely right. There is indeed such a thing called ‘definite integral’.
We will cover ‘definite integrals’ in a couple of lectures. ‘Indefinite integrals’ and
‘definite integrals’ are related, but today I am not going to discuss ‘definite integrals’
at all. I will eventually cover ‘definite integrals’ in due course. So, look forward to it.
• Find indefinite integrals.
So far what we have covered: You are given a concrete
f
x
. You are also given
a concrete F x . You are told the fact that F x is an antiderivative of f x .
Then you have practiced how to write the same fact using the symbol
Z
.
20
This was easy. But you must have certainly expected to see questions in the following
format, in the assignment and in the tests:
Format. You are given a concrete f x , but you are not given its antiderivative F x . Find
Z
f x dx.
Questions in this exact format is what we are going to practice next. We are going
to do a few dozens, until you get tired of it. Please don’t say “oh, that is too much”.
This is a necessary part.
• Before throwing actual problems, I give you some tips. One of the tips is the
following rule, called “Break-up Rule”. This is the indefinite integral version of the
rule about differentiation you have learned under the same name. Like the derivative
break-up rule, it comes with three parts.
• Break-up Rule.
Suppose
is an antiderivative of f x .
G x is an antiderivative of g x .
F x
Then
(i)
Z
f x
+ g x
dx
=
F x + G x + C.
(ii)
Z
f x
− g x
dx
=
F x − G x + C.
(iii)
Z
a f x dx
=
a F x
+ C
21
a is a constant .
• Here is another rule that comes handy:
Rule (x-to-the-constant-power integration rule).
Z
xα dx
1
α+1
=
The following compensates the above
xα+1 + C
α is a constant ;
it takes care of the case α = −1 :
Rule (x-to-the-(−1)-power integration rule).
Z
x−1 dx
=
ln x + C
.
Z
1
dx
x
=
ln x + C
.
Paphrase:
Practice problems – The indefinite integrals.
Z
(1)
(2)
Z
α 6= −1 .
6 x3 dx.
1 7
x dx.
3
(3)
Z 3 2
x + 3 dx.
4
(4)
Z 8 5
x + 10 x4 dx.
5
22
Z (5)
Z (6)
(7)
Z x−1
x+1
x+1
x+3
2
x +x+1
(8)
(9)
Z
Z (10)
(11)
Z x+
1 x−
x
6
x2
(14)
=
=
4
x5
2
x −1
2
dx
x + 4x + 3
.
dx
.
−
dx.
3
dx.
2 x3
Z
4
dx.
x
Z
Z 2 x−3 dx.
Z −2
x
dx.
Z 1 dx
x
(12)
(13)
dx
dx
Z
+
=
1 dx.
x3
5 √
x dx.
4
23
Z 2
x −
1 dx .
x2
Z
(15)
Z
5
x−11/3 dx.
3
dx
dx
(16)
Z
(17)
(18)
Z
x
1
√
x 2x
√
x +2
7 x5/2 dx.
(19)
Z
(20)
Z √
(21)
Z
4x+1
e
Z
(22)
(23)
(24)
dx
Z
=
Z =
x
x
3/2
−3/2
2 e−2x dx.
Z
=
4x
e e
2
3 x ex dx.
4
x3 e−x dx.
ln x
2x
24
dx.
dx
+ 2x
5 ex dx.
1
√
2
Z
1
√
2
Z
dx
.
.
dx
.
(25)
Z
(26)
Z
9
(28)
(29)
Z
(31)
(33)
x+1
Z
(30)
(32)
2
Z
5
x
6
Z
x
4
dx
2x
dx
1
2x
dx
+ 1
x
x
8
x
dx
2 e dx
dx.
dx.
3 + ln x
x ln x
Z
Z
ln x
2
x ln x
Z
(27)
dx.
2 · 3x dx.
=
=
x
2 · 2 dx
Z
x
16 dx
.
.
Z 1 x
dx .
2
=
Z
=
=
25
Z Z x
x
x
30 + 5
2e
dx
.
dx
.
Remark (Cautionary examples).
Finding antiderivatives and finding derivatives are completely different ballgames. Here are some cautionary examples.
(1)
We know the fact that the derivative of
x
ln x
is
1
ln x
1
−
ln x
2 .
To write the same fact in one line:
d x
dx
ln x
1
ln x
=
1
−
ln x
x
,
ln x
You can just physically perform differentiation of
2 .
and will come to the
same conclusion
as mine. Please do not skip this — it’s an easy application
of the
‘quotient rule’. Now, here we go: A paraphrase of the same fact as usual :
Paraphrase.
x
ln x
is an antiderivative of
1
ln x
1
−
ln x
2 .
dx
=
Namely,
Z 1
ln x
−
1
ln x
2
26
x
ln x
+ C.
• Now, having said, the subtlety of the matter is, the above does not provide us
information as to how to find an antiderivative of
of
1
ln x
2 .
1
,
ln x
or an antiderivative
So, if you ask how much is
1
ln x
Z
dx,
or how much is
Z
1
ln x
2
dx,
I have no idea. Nobody has any idea.4 You would say “What?”
The truth is, there is no way to express those . You cannot write
Z
1
ln x
dx
Z
or
1
ln x
2
dx
using known “arguments and formations” as a quantity of x. Here, known “arguments and formations” meansaddition, subtraction, multiplication, reciprocation, ex-
ponentiation raising powers , radicals such as square-rooting, cube-rooting, etc.
and the logarithm.
Some literatures use the term “elementary functions” in reference to quantities
involving x obtained through known “arguments and formations”. If you allow me
to use this terminology, then what I just said above is paraphrased as follows: Neither
of
Z
Z
1
1
dx
nor
2 dx
ln x
ln x
is an ‘elementary function’.
4 The
mathematicians community reached the following agreement: An antiderivative of 1 ln x
is always denoted Li x.
27
1
ln x
You agree that both
1
and
ln x
are elementary functions,
2
because these are evidently constructed from x through the known “arguments and
formations”. What I am saying, therefore, is summarized in oneline:
“Antiderivatives of elementary functions are not always elementary.”
This sounds a little confusing, and you probably want me to elaborate this. This
is actually a subtle, and ‘intense’, topic. Today somehow I do not want to delve
into this. I am going to revisit this issue in “Review of Lectures – VIII”. So far, all
you need to know is just to understand that antiderivative problems are not always
“doable”, or “feasible”.
(2) Here is another example of an antiderivative problem that is not “doable” or
2
“feasible”: An antiderivative of ex
is not an elementary function. Similarly, an
−x2
is not an elementary function.
antiderivative of e
In the plain language, you cannot write either of
Z
x2
e
dx
or
Z
2
e−x dx
using known “arguments and formations” as a quantity of x.
Analyzing these antiderivatives is actually not easy. But as I said, today I do
not want to go any further. I just want you to agree that antiderivative problems
are ‘non-trivial’ affairs. We will come back to this topic in “Review of Lectures –
VIII”. Aren’t you curious about this aspect of integrals? Can I make a subjective
statement? Calculus is not a dull subject, but rather it is a very stimulating subject.
Finally, one more exercise.
28
Exercise 14.
Verify:
(1)
F x
= −
ln x + 1
x
(2)
F x
= −
2 ln x + 1
4 x2
is an antiderivative of f x
=
ln x
.
x3
(3)
F x
= −
3 ln x + 1
9 x3
is an antiderivative of f x
=
ln x
.
x4
(4)
F x
= x
(5)
F x
=
x
is an antiderivative of f x
is an antiderivative of f x
e−x
x
=
ln x
is an antiderivative of
f x
= −
1
x
29
+
1
x2
e−x .
ln x
.
x2
=
+ 1
xx .
(1) through (33), pages 22 through 25 .
Solution for Practice problems
Z
(1)
(4)
(5)
Z (6)
(7)
x2 + x + 1
Z x−1
x+1
dx
=
x+1
x+3
dx
1 3
x + 3 x + C.
4
=
Z 8 5
4
dx
x + 10 x
5
Z 1
x8 + C.
24
=
Z 3 2
x + 3 dx
4
(3)
3 4
x + C.
2
=
1 7
x dx
3
Z
(2)
6 x3 dx
4
x6 + 2 x5 + C.
15
=
1 3
x +
3
1 3
x − x + C.
3
=
dx
1 2
x + x + C.
2
1 3
x + 2 x2 + 3 x + C.
3
=
(8)
Z
2 x−3 dx
=
−1
x2
+ C.
(9)
Z
4
dx
x5
=
−1
x4
+ C.
(10)
(11)
Z Z 6
x2
x+
−
3
dx
2 x3
1 x−
x
−6
x
=
1 dx
x
30
=
+
3
4 x2
1 3
x +
3
+ C.
1
x
+ C.
4
dx
x
=
4 ln x + C.
1 dx
x3
=
−2 ln x −
5 √
x dx
4
=
7 x5/2 dx
=
5
x−11/3 dx
3
=
Z
(12)
(13)
Z −2
x
+
Z
(14)
Z
(15)
Z
(16)
(17)
(18)
1
√
x 2x
Z
Z
x
√
x +2
Z
(19)
dx
(20)
2 e−2x dx
(21)
Z
e4x+1 dx
(22)
=
5 ex dx
Z √
Z
=
dx
2
3 x ex dx
=
=
1
2 x2
5
x3/2 + C.
6
2 x7/2 + C.
−5 −8/3
x
+ C.
8
√
− 2
√
x
+ C.
2 5/2
x
+ x2 + C.
5
5 ex + C.
−1 −2x
√
e
+ C.
2
1 4x+1
e
+ C.
4
=
=
31
+ C.
3 x2
e
+ C.
2
1
√
2
Z
(23)
4
x3 e−x dx
ln x
2x
Z
(24)
(25)
Z
(26)
Z
9
ln x
x
Z
(28)
(29)
2x+1 dx
(30)
Z
42x dx
(31)
(32)
(33)
1
2x
Z
Z
5x
6x + 1
Z
x
dx
=
ln x
2 ln
ln x
ln x
=
=
2 · 3x dx
Z
x
dx
2 e dx
3 ln
2
ln 3
=
2
ln 2
=
=
1
ln 16
+ C.
+ C.
+ ln x + C.
3x + C.
2x + C.
1
2x
=
1
ln 30
=
1
1 + ln 2
32
9
16x + C.
−1
ln 2
=
dx
+ C.
2
1 ln x
+ C.
4
dx
dx
4
e−x
=
8
2
x ln x
3 + ln x
x ln x
Z
(27)
dx
−1
√
4 2
=
+ C.
1
ln 5
30x +
2e
x
5x + C.
+ C.