mat01a1: antiderivatives · are also antiderivatives of f(x) = 2xbecause g0(x) = 2x and h0(x) = 2x....

MAT01A1: Antiderivatives

Dr Craig

Week: 11 May 2020

Warm up

Calculate the limit: limx→−4

sin(πx)

x2 − 16

Attempt this on your own before looking at

the solution on the next slide.

limx→−4

sin(πx)

x2 − 16We have lim

x→−4sin(πx) = sin(−4π) = 0 and

limx→−4

(x2 − 16) = (−4)2 − 16 = 0, so we

have a 00 indeterminate form.

limx→−4

sin(πx)

x2 − 16

(0

0

)L’H= lim

x→−4

π cos(πx)

2x

=π cos(−4π)

2(−4)=

π

−8≈ −0.393

Warm up

Calculate the limit: limx→−4

sin(πx)

x2 − 16

An ∞−∞ indeterminate form example

limx→∞

(lnx− x)

Attempt this before looking at the solution

on the next slide.

Hint: take out a “common factor”.

limx→∞

(lnx− x)

Solution:

limx→∞

(lnx− x) = limx→∞

x

(lnx

x− 1

)We want to see what happens to lnx

x as

x→∞.

limx→∞

lnx

x

(∞∞

)L’H= lim

x→∞

1/x

1= lim

x→∞

1

x= 0

limx→∞

(lnx− x)Solution:

limx→∞


x

(lnx

x− 1

)

We want to see what happens to lnxx as

x→∞.

limx→∞

lnx

x

(∞∞

)L’H= lim

x→∞

1/x

1= lim

x→∞

1

x= 0

limx→∞

(lnx− x)Solution:

limx→∞


x

(lnx

x− 1

)We want to see what happens to lnx

x as

x→∞.

limx→∞

lnx

x

(∞∞

)L’H= lim

x→∞

1/x

1= lim

x→∞

1

x= 0

limx→∞

(lnx− x)

Solution continued. . . We have

limx→∞

lnx

x= 0 so

limx→∞

(lnx

x− 1

)= −1.

Therefore

limx→∞

x

(lnx

x− 1

)= −∞

After that brief warm-up, we move on to the

topic of Ch 4.9: antiderivatives. You can

probably already guess what is contained in

this section.

Chapter 3 was all about how to go from a

function f to its derivative f ′. This section

is about going in the opposite direction.

That is, if we are given some function f ,

what is the function F such that F ′ = f .

Antiderivatives

A function F is an antiderivative of f on

an interval I if

F ′(x) = f (x)

for all x ∈ I .

Example: F (x) = x2 is an antiderivative of

f (x) = 2x.

Question: are antiderivatives unique?

Answer: No, antiderivatives are not unique.

Consider the previous example of f (x) = 2x

and F (x) = x2. The functions

G(x) = x2 + 4 or H(x) = x2 + 7

are also antiderivatives of f (x) = 2x because

G′(x) = 2x and H ′(x) = 2x



and F (x) = x2.

The functions

G(x) = x2 + 4 or H(x) = x2 + 7


G′(x) = 2x and H ′(x) = 2x



and F (x) = x2. The functions

G(x) = x2 + 4 or H(x) = x2 + 7


G′(x) = 2x and H ′(x) = 2x

Theorem: If F is an antiderivative of f

on an interval I , then the most general

antiderivative of f on I is F (x) + C

where C is an arbitrary constant.

Examples of general antiderivatives

Find the general antiderivative of

f (x) = 3x2 + 2x + 6

Answer: F (x) = x3 + x2 + 6x + C

Harder: find the general antiderivative of

g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3 +

3

2x2 + 4x + C



f (x) = 3x2 + 2x + 6

Answer: F (x) = x3

+ x2 + 6x + C


g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3 +

3

2x2 + 4x + C



f (x) = 3x2 + 2x + 6

Answer: F (x) = x3 + x2

+ 6x + C


g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3 +

3

2x2 + 4x + C



f (x) = 3x2 + 2x + 6

Answer: F (x) = x3 + x2 + 6x

+ C


g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3 +

3

2x2 + 4x + C



f (x) = 3x2 + 2x + 6

Answer: F (x) = x3 + x2 + 6x + C


g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3 +

3

2x2 + 4x + C



f (x) = 3x2 + 2x + 6

Answer: F (x) = x3 + x2 + 6x + C


g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3

+3

2x2 + 4x + C



f (x) = 3x2 + 2x + 6

Answer: F (x) = x3 + x2 + 6x + C


g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3 +

3

2x2

+ 4x + C



f (x) = 3x2 + 2x + 6

Answer: F (x) = x3 + x2 + 6x + C


g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3 +

3

2x2 + 4x

+ C



f (x) = 3x2 + 2x + 6

Answer: F (x) = x3 + x2 + 6x + C


g(x) = 7x2 + 3x + 4

Answer: G(x) =7

3x3 +

3

2x2 + 4x + C

More examples:


f (x) = sinx

Answer: F (x) = − cosx + C

Function Particular antiderivative

cf (x) cF (x)

f (x) + g(x) F (x) +G(x)

xn (n 6= −1) 1n+1x

n+1

1

x`n|x|

ex ex

Function Particular antiderivative

cosx sinx

sinx − cosx

sec2 x tanx

secx tanx secx

Function Particular antiderivative1√

1− x2arcsinx

1

1 + x2arctanx

coshx sinhx

sinhx coshx

In the examples on this slide, we are not

looking for the most general antiderivative,

but for the antiderivative that satisfies some

additional conditions.

Examples:

1. Find f if f ′′(x) = 12x2 + 6x− 4,

f (0) = 4 and f (1) = 1.

2. Find f if f ′(x) = ex + 20(1 + x2)−1 and

f (0) = −2.

Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4

and f (1) = 1.

Solution (1.):

Using the reverse of the power

rule we get f ′(x) = 4x3 + 3x2 − 4x + C.

This gives f (x)=x4 + x3 − 2x2 + Cx +D.

Now we substitute x = 0 and get

4 = 04 + 03 − 2(02) + C(0) +D,

so D = 4. Now substitute x = 1 and get

1 = 14 + 13 − 2(12) + C(1) + 4,

so C = −3. Our final function is

f (x) = x4 + x3 − 2x2 − 3x + 4.

Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4

and f (1) = 1.

Solution (1.): Using the reverse of the power




4 = 04 + 03 − 2(02) + C(0) +D,


1 = 14 + 13 − 2(12) + C(1) + 4,


f (x) = x4 + x3 − 2x2 − 3x + 4.

Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4

and f (1) = 1.





4 = 04 + 03 − 2(02) + C(0) +D,

so D = 4.

Now substitute x = 1 and get

1 = 14 + 13 − 2(12) + C(1) + 4,


f (x) = x4 + x3 − 2x2 − 3x + 4.

Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4

and f (1) = 1.





4 = 04 + 03 − 2(02) + C(0) +D,


1 = 14 + 13 − 2(12) + C(1) + 4,


f (x) = x4 + x3 − 2x2 − 3x + 4.

Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4

and f (1) = 1.





4 = 04 + 03 − 2(02) + C(0) +D,


1 = 14 + 13 − 2(12) + C(1) + 4,

so C = −3.

Our final function is

f (x) = x4 + x3 − 2x2 − 3x + 4.

Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4

and f (1) = 1.





4 = 04 + 03 − 2(02) + C(0) +D,


1 = 14 + 13 − 2(12) + C(1) + 4,


f (x) = x4 + x3 − 2x2 − 3x + 4.

Find f if f ′(x) = ex + 20(1 + x2)−1 and

f (0) = −2.

Solution (2.):

Recall that

d

dx(arctanx) =

1

1 + x2.

Therefore f (x) = ex + 20 arctanx + C. We

substitute x = 0 and get

−2 = e0+20 arctan 0+C =⇒ C = −3.

Therefore f (x) = ex + 20 arctan(x)− 3


f (0) = −2.

Solution (2.): Recall that

d

dx(arctanx) =

1

1 + x2.



−2 = e0+20 arctan 0+C =⇒ C = −3.



f (0) = −2.


d

dx(arctanx) =

1

1 + x2.

Therefore f (x) = ex + 20 arctanx + C.

We


−2 = e0+20 arctan 0+C =⇒ C = −3.



f (0) = −2.


d

dx(arctanx) =

1

1 + x2.



−2 = e0+20 arctan 0+C

=⇒ C = −3.



f (0) = −2.


d

dx(arctanx) =

1

1 + x2.



−2 = e0+20 arctan 0+C =⇒ C = −3.


REMEMBER THE CHAIN RULE!

Find the most general antiderivative of:

f (x) = cosx. sec2(sinx)

We will only learn a formal method for this in

Ch 5.5 but it is worth introducing the idea

already now. Notice that cosx is the

derivative of sinx which is the inner function

of the composite function sec2(sinx).

Looking at this (and thinking hard) we get:

F (x) = tan(sinx) + C






already now.

Notice that cosx is the










already now. Notice that cosx is the





Below is another example of this type.

Find the most general antiderivative of

f (x) = 3x2ex3

Notice here that ddx(x

3) = 3x2. From this we

get:

F (x) = ex3+ C

As we mentioned, we will introduce a formal

technique for dealing with antiderivatives of

this type a bit later on.



f (x) = 3x2ex3


3) = 3x2.

From this we

get:

F (x) = ex3+ C






f (x) = 3x2ex3


3) = 3x2. From this we

get:

F (x) = ex3+ C




Prescribed tut problems:

Complete the following exercises from the

8th edition (Ch 4.9):

1, 3, 5, 11, 13, 15, 17, 25, 29, 37, 41

mat01a1: antiderivatives · are also antiderivatives of f(x) = 2xbecause g0(x) = 2x and h0(x) = 2x....

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