SPECIALIST MATHS

Calculus

Week 6

Definite Integrals & Areas

Definite Integrals

• If is a continuous function for

• and

• The Fundamental Theorem of Calculus

xf

dxxfxFbxa

b

aaFbFdxxf

What happened to the constant c

b

acaFcbFdxxf

b

acaFcbFdxxf

b

aaFbFdxxf

Definitions

• The function is called the integrand

• and are called the lower and upper bounds

• is the indefinite integral

• is called the definite integral

b

aaFbFdxxf

b

adxxf

a b

xf

dxxf

Properties of Definite Integrals

a

adxxf

0

b

a

a

bdxxfdxxf

b

a

b

adxxfkdxxkf

dxxgdxxfdxxgxfb

a

b

a

b

a

)()]([

dxxgdxxfdxxfb

m

b

a

m

a

)(

Example 22 Evaluate 3

0 2cos

dxx

Solution 22Evaluate 3

0 2cos

dxx

3

0 2cos

dxx

02sin

2

1

3

2sin

2

1

02

3

2

1

4

3

3

0

2sin2

1

x

Graphics Calculator for Example 1

3

0 2cos

dxx

RUN

OPTN F4 (calc) )(dx

)3,0),2(cos( X

F4

SHIFT SETUP Arrow down to ANGLE

F2(Rad) EXIT

Ans = 0.4330

(CASSIO)

Example 23 6

12 3cot :form surdin Evaluate

dxx

Solution 23

6

12 3cot

dxx

6

12 3sin

3cos

dxx

x

6

12 3cos

3sin

1

xdx

x

Solution 23 (cont) 6

12 3cos

3sin

1

xdx

x

xu 3sin

xdx

du3cos3

dx

dux

3

13cos

1 2

sin6

3sin)

6(

u 2

1

4sin

12

3sin

12

u

6

12 3cos

3sin

1

xdx

x 1

21 3

11 dx

dx

du

u

Solution 23 (continued)

6

12 3cos

3sin

1

xdx

x

1

21 3

11 dx

dx

du

u

1

21

1

3

1duu1

21

ln3

1

u

2

1ln

3

11ln

3

1

2

12ln

3

10

2ln2

1

3

1

2ln6

1

Area Determination• If is a positive continuous function in

the interval then the shaded area is given by

)(xf

bxa

b

axf

dx )(

a b

y

x

)(xfy

Area for function that is also negative

a b c

y

x

c

b

b

adxxfdxxf

)( )(Area

Example 24

.x0for axis -x

theand4sin2x curve ebetween th area theFind

y

Solution 24• Find the area between the curve

and the x-axis forxy 2sin4

x0

Step 1 draw the curve on graphics calculator and find the x intercepts

y

xo2

xxx

x

kxkx

kxkx

x

x

or 2

,0

0for 2

or 0

22or 202

02sin

02sin4

Solution 24 continuedStep 2 state the area relation

2

2

0 dx 2sin4dx 2sin4Area xx

Step 3 calculate definite integral either with G Calc or algebraically

2

2

0

2cos2

42cos

2

4

xx

2 2

0 2cos22cos2 xx cos22cos20cos2cos2

12121212

44 2units 8

Area between two curvesy

y

)(xfy

)(xgy

a b

b

adxxgxfArea

)()(

Example 25

x

xyxy

0for

2cos and cosbetween area theFind

Solution 25• Find the area between

xxyxy 0for 2cos and cosStep 1 draw the graph.

xy cos

xy 2cosy

x

Solution 25 continuedStep 2 find the points of intersection

xx 2coscos

1cos2cos 2 xx

1coscos20 2 xx

)1)(cos1cos2(0 xx

1cosor 2

1cos xx

0or 3

2 xx

Solution 25 continuedStep 3 state the definite integral for the area and calculate it

32

0 2coscosArea

dxxx

32

0

2sin2

1sin

xx

0sin

2

10sin

3

4sin

2

1

3

2sin

2

3

2

1

2

3

4

3

2

3

4

332

2units 4

33

This Week• Text Book Pages 258 to 263

• Exercise 7D3 Q 1 – 2

• Exercise 7D4 Q 1 – 6

• Exercise 7D5 Q 1 – 3

• Questions 5 & 6 from Review Sets 6A – 6C

• Review Sets 7A – 7D