Lesson 13

Solving Definite Integrals

How to find antiderivatives

We have three methods:

1. Basic formulas

2. Algebraic simplification

3. Substitution

Basic Formulas

If f(x) is… …then an antiderivative is…

�

xn

�

1

n+1xn+1 except if n = –1

k kx (assuming the variable is x!)cos(kx) sin(kx)/ksin(kx) –cos(kx)/kekx ekx /k

�

1

x

�

ln x

ax ax /ln(a)

Algebraic Simplification2 31

3( 1)( 1) 1x x dx x dx x x C! + = ! = ! +" "

212

(1 )1

x xdx x dx x x C

x

+= + = + +! !

2x x

dxx

+=!

2 3 2 3 29 6

3 29 6 1 3 3x x dx x x x C x x x C+ + = + + + = + + +!

( )2

3 1x dx+ =!

Integrals by SubstitutionStart with

Let u = g(x).du

�

du

dx= ! g (x) so

�

du = ! g (x)dx

Let u =?

�

x sin x2( )! dx

2 2 21 1 1sin( ) sin( )2 sin( ) cos( )

2 2 2x x dx x x dx u du x C= = = ! +" " "

u = x2 ! du = 2xdx

Problem 1 Use basic formulas:

Antiderivative Practice

�

2sin(3t) ! e4 t + 4tdt"

�

z2

+ 2z

z2! dz

�

6t

4 + t2! dt

Problem 2 Simplify algebraically first, then integrate.

Problem 3 Make a substitution: Let u=4+t2, so du=2tdt.

2 2

2 2 2

2 2 21 2ln

z z z zdz dz dz z z C

z z z z

+= + = + = + +! ! !

2

2 2

6 2 13 3 3ln 3ln 4

4 4

t tdt dt du u C t C

t t u= = = + = + +

+ +! ! !

2 sin(3t) ! e4 t + 4 t dt" = ! 23 cos(3t) !

14 e

4 t+

1ln(4) 4

t+ C

Antiderivative Practice

�

2

y ln(ky)dy!

Find the particular function F(x) such that F'(x) = x2 and the graph of F(x) passes through (1, 2).

Problem 4

Problem 5

Make a substitution: u=ln(ky), so du=dy/y.

2 2 1 22ln 2ln ln( )

ln( ) ln( )dy dy du u C ky C

y ky ky y u= = = + = +! ! !

The general antiderivative is

�

x2! dx =

1

3x3

+ C

Then to find C, we must have

�

F(1) =1

313

+ C = 2

Thus, C = 5/3, and our function is

�

F(x) =1

3x3

+5

3

Solving Definite Integrals

Theorem: (Fundamental Theorem I)

Or: If F is an antiderivative for f, then

We determined using Simpson’s Rule :

Now use the fundamental theorem:

An antiderivative for f(x) = 3x + 5 is

So:12

03 5 F(12) - F(0) = 276 - 0 = 276x dx+ =!

Example

Example

We have to• find an antiderivative;• evaluate at 3;• evaluate at 2;• subtract the results.

�

x3dx =

2

3

! 1

4x4

2

3

=1

434 " 1

424

=81

4" 16

4=65

4=16.25

33

2

x dx!

This notation means:evaluate the function at3 and 2, and subtract the

results.

Don’t need to include “+ C” in ourantiderivative, because anyantiderivative will work.

Examples02sin( ) 3x x dx

!

+"

( )

( )

2 2

00

2 2

3 32sin( ) 3 2cos 2cos 2cos0 0

2 2

3 3 = 2( 1) 2(1) 0 4

2 2

xx x dx x

!! !

!

! !

" #+ = $ + = $ + $ $ +% &

' (

" #$ $ + $ $ + = +% &

' (

)

Alternate notation

= –1 = 1

1

2

1ds

s

!

!"1

s!2

!1

" ds = ln s!2

!1= ln1! ln2 =!! ln2

Practice Examples

ses+1

s!2

!1

" ds

= es+1

s!2

!1

" ds = (es+ ln s )

!2

!1= e

!1+ ln1! (e!2 + ln2)

=!!1

e!1

e2! ln2

�

3 s ds2

9

! = 3s

1/ 2ds

2

9

! = 2s3/ 2

2

9

= 2 9( )3/ 2

" 2 2( )3/ 2

= 54 " 4 2

1

edx

1

5

!

=4

e

Substitution in Definite Integrals• We can use substitution in definite integrals.• However, the limits are in terms of the original variable.• We get two approaches:

– Solve an indefinite integral first– Change the limits

First solve an indefinite integral to find an antiderivative.

Then use that antiderivative to solve the definite integral.

Note: Do not say that a definite and an indefinite integral are equalto each other! They can’t be.

Method I:

Example

First: Solve an indefinite integral.u = t2 + 4

du = 2t dt

�

3t

t2

+ 4

! dt

�

3t

t2

+ 4dt

! =3

2

2t

t2

+ 4dt

! =3

2

1

udu

! =3

2ln t

2+ 4 + C

3t

t2+ 41

2

! dt = 3

2ln t

2+ 4( )

1

2

= 3

2ln 8( ) " 3

2ln 5( ) = 3

2ln

8

5

Pull out the 3,and put in a 2.

2t dtbecomes du

Here’s anantiderivative!

Second: Use the antiderivative to solve the definite integral.

Here’s the antiderivativewe just found.

ExampleWhen discussing population growth, we worked backwards to find outwhat we got from evaluating

Let’s find an antiderivative using substitution.

�

! P ( t)

P( t)" dt =

1

udu = ln u + C = ln P(t) + C"

�

! P ( t)

P(t)dt

a

b" = ln P( t)( )a

b

= ln P(b)( )# ln P(a)( ) = lnP(b)

P(a)

$

% &

'

( )

u = P(t)

du = P'(t) dt( )

( )

P tdt

P t

!"

Of course, P(t) is always non-negative, so we don’t need absolutevalues…

So we get:

Method II: Convert the LimitsWe start with x = a and x = b, and a substitution formula u = …

Just put a and b into the substitution formula and get new limits.

Note: You do not have to go back to x then!

�

3t

t2

+ 41

2

! dt

2 2

2 21 1

3 3 12

4 2 4

tdt t dt

t t=

+ +! !

Example Start with the same substitutionu = t2 + 4

du = 2t dt

becomes u becomes du

( )8 8

3 3 8

2 2 555

3 1ln ln

2du uu

= = =!

What happens to t = 1?

u = t2 + 4 = 1^2 + 4 = 5.And when t = 2,u =  t2 + 4 = 2^2 + 4 = 8.

When t = 1, u = 5.

When t = 2, u = 8.

Example

( ) ( ) ( )

320

4

11 1 38 8 8 3202 2 23 33 3

1 1 1 5

5

4 4

3 3

1 1 15 5 5 10

410 10 10

3

3 320 5 163.5

40

uy y dy y y dy y y dy u du= = = =

! "= # $% &

' (

) ) ) )

( )8

23

1

5y y dy!u = 5y2

du = 10y dy

y = 1: u = 5

y = 8: u = 320

y 5y23( )1

8

! dy = y 5y2( )1

3

1

8

! dy = 5

1

3 y

2

3"

#$%

&'1

8

! y dy = 5

1

3 y

5

3

1

8

! dy

= 5

1

3y

5

3+1

5

3+1

1

8

= 5

1

33

8(8

8

3 (1

8

3 ) ) 163.5

Note that we can also dothis problem without u-sub--try algebraic simplification

Practice Example

e4x

1+ e4x0

1

! dx

e4x

1+ e4x

dx!Method I: Firstly compute

u = 1+ e4x

du = 4e4x

dx

e4x

1+ e4x

dx! =1

4

4e4x

1+ e4x

dx! =1

4

1

u

du =!1

4u"

1

2 du =!1

4

u"

1

2+1

"1

2+1

+ C

=1

2u

1

2 + C =1+ e

4x

2+ C

e4x

1+ e4x0

1

! dx =1

21+ e

4x |0

1=

1

21+ e

4 "1

21+ e

0=

1

21+ e

4 "1

22

Method II:

e4x

1+ e4x

dx0

1

! =1

4

4e4x

1+ e4x

dx0

1

! =1

4

1

u

du =2

1+e4

!1

4u"

1

2 du =2

1+e4

!1

4

u"

1

2+1

"1

2+1

|2

1+e4

=1

2u

1

2 |2

1+e4

=1+ e

4

2"

2

2

u = 1+ e4x

u(0) = 1+ e4*0

= 2,u(1) = 1+ e4*1

= 1+ e4

Using Definite IntegralsWe can now evaluate many of the integrals that we have been ableto set up.

Find area between y = sin(x) and thex–axis from x = 0 to x = π, and fromx = 0 to x = 2π.

The area from 0 to π is clearly:

Example

0 0sin( ) cos( ) 1 1 2x dx x

!!

= " = + =#The area from 0 to 2π is more complicated. We note that

But this is obviously not the area!

2

0sin( ) 0x dx

!

="

The area from 0 to 2π can be found by:

( ) ( )2 2

00sin( ) sin( ) cos( ) cos( ) 4x dx x dx x x

! ! ! !

!!" = " + =# #

Summary

• Used the fundamental theorem to evaluatedefinite integrals.

• Made substitutions in definite integrals– By solving an indefinite integral first– By changing the limits

• Used the fundamental theorem to evaluateintegrals which come from applications.