1

Chapter 7Integral Calculus

The basic concepts of differential calculus were covered in the preceding chapter. This chapter will be devoted to integral calculus, which is the other broad area of calculus. The next chapter will be devoted to how both differential and integral calculus manipulations can be performed with MATLAB.

2

Anti-Derivatives

An anti-derivative of a function f(x) is a new function F(x) such that

( )( )

dF xf x

dx

3

Indefinite and Definite Integrals

( )f x dx2

1

( )x

xf x dx

Indefinite

Definite

4

Definite Integral as Area Under the Curve

Approximate Area kk

y x

1y 2y3y

4y

Ky

a b x

y

5y

5

Exact Area as Definite Integral

limb

ka x dxk

ydx y x

6

Definite Integral with Variable Upper Limit

x

aydx

( )x

ay u du

More “proper” form with “dummy” variable

7

Area Under a Straight-Line Segment

1x 2x x

1y

2y( )y f x

L

2

12 1 2 1

1

2

x

xydx y y x x

8

Example 7-1. Determine12

0ydx

20 20

-20

-10

24 6

89 12

x

( )y f x

9

Example 7-1. Continuation.

12

040 0 20 20 0 30 10ydx

20 20

-20

-10

24 6

89 12

x

( )y f x

10

Example 7-2. Determine0

xydx

20 20

-20

-10

24 6

89 12

x

( )y f x

11

Guidelines

1. If y is a non-zero constant, integral is either increasing or decreasing linearly.

2. If segment is triangular, integral is increasing or decreasing as a parabola.

3. If y=0, integral remains at previous level.4. Integral moves up or down from previous

level; i.e., no sudden jumps.5. Beginning and end points are good

reference levels.

12

20 20

-20

-10

24 6

89 12

x

( )y f x(a)

40

20

40

10

x2 4 6 8 9 12

0 0( )

x xydx f x dx (b)

13

Tabulation of Integrals

( ) ( )F x f x dx( )

b

aI f x dx

( ) ( ) ( )b

aI F x F b F a

14

Table 7-1. Common Integrals.( )f x ( ) ( )F x f x d x I n t e g r a l N u m b e r

( )a f x ( )a F x I - 1

( ) ( )u x v x ( ) ( )u x d x v x d x I - 2

a a x I - 3

1nx n 1

1

nx

n

I - 4

a xe a xe

a

I - 5

1

x l n x I - 6

s i n a x1

c o s a xa

I - 7

c o s a x1

s i n a xa

I - 8

2s i n a x1 1

s i n 22 4x a x

a I - 9

15

Table 7-1. Continuation.

2c o s a x1 1

s in 22 4x a x

a I -1 0

s inx a x 2

1s in c o s

xa x a x

a a I -1 1

c o sx a x 2

1c o s s in

xa x a x

a a I -1 2

s in c o sa x a x 21s in

2a x

aI -1 3

s in c o sa x b x2 2fo r a b

c o s ( ) c o s ( )

2 ( ) 2 ( )

a b x a b x

a b a b

I -1 4

a xx e 21

a xea x

a I -1 5

ln x ln 1x x I -1 6

2

1

a x b11

ta na

xba b

I -1 7

16

In Examples 7-3 through 7-5 that follow, determine the following integral in each case:

z ydx

17

Example 7-3

412 xy e4

4

4

12 124

3

xx

x

ez e dx C

e C

18

Example 7-4

12 sin 2y x x

2

12 sin 2

112 sin 2 cos 2

(2) 2

3sin 2 6 cos 2

z x xdx

xx x C

x x x C

19

Example 7-5

2 36y x

x

2

2

3

3

36

36

63ln

3

2 3ln

z x dxx

x dx dxx

xx C

x x C

20

In Examples 7-6 and 7-7 that follow, determine the definite integral in each case as defined below.

b

aI ydx

21

Example 7-6

0sinI xdx

00sin cos

cos cos 0

( 1) ( 1) 2

I xdx x

22

Example 7-7

1 2

08 xI xe dx

1 2

0

21

2 0

2 0

2

8

8 2 1(2)

2 2(1) 1 2 0 1

6 2 1.188

x

x

I xe dx

ex

e e

e

23

Displacement, Velocity, and Acceleration

2 2( ) acceleration in meters/second (m/s )

( ) velocity in meters/second (m/s)

( ) displacement in meters m

a a t

v v t

y y t

( )dv

a tdt

( )dv

dv dt a t dtdt

( )dv a t dt dv v

24

Displacement, Velocity, and AccelerationContinuation

1( )v a t dt C ( )

dyv t

dt ( )

dydy dt v t dt

dt

2( )y v t dt C

25

Alternate Formulation in Terms of Definite Integrals

0( ) ( ) (0)

tv t a t dt v

0( ) ( ) (0)

ty t v t dt y

26

Example 7-8. An object experiences acceleration as given by

2( ) 20 ta t e

Determine the velocity and displacement.

(0) 0

(0) 0

v

y

27

Example 7-8. Continuation.

2( ) 20 tdva t e

dt

2 2 21 1

20( ) 20 10

2t t tv t e dt e C e C

0

1 1(0) 10 10 0v e C C

1 10C 2( ) 10 10 tv t e

28

Example 7-8. Continuation.

2

2 22 2

( ) 10 10

1010 10 5

2

t

t t

y t e dt

t e C t e C

02 2(0) 0 5 5 0y e C C

2 5C

( ) 10 5 5ty t t e

29

Example 7-9. Rework previous example using definite integral forms.

2 2

0 00

2 0 2

20( ) ( ) (0) 20 0

2

10 10 10 10

tt t t t

t t

v t a t dt v e dt e

e e e

2 2

0 0 0

2 0 2

( ) ( ) (0) (10 10 ) 10 5

10 5 0 5 10 5 5

tt t t t

t t

y t v t dt y e dt t e

t e e t e

30

Example 7-10. Plot the three functions of the preceding examples.

2( ) 20 ta t e2( ) 10 10 tv t e

2( ) 10 5 5ty t t e

31

Example 7-10. Continuation.

>> t = 0:0.02:2;>> a = 20*exp(-2*t);>> v = 10 -10*exp(-2*t);>> y = 10*t + 5*exp(-2*t) - 5;>> plot(t, a, t, v, t, y)

The plots are shown on the next slide.

32