25: definite integration © christine crisp “teach a level maths” vol. 1: as core modules

25: Definite 25: Definite IntegrationIntegration

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

Definite Integration

The numbers on the integral sign are called the limits of integration

e.g.1

1

2

2 23 dxx is a definite integral


xx

23

3 3

Find the indefinite integral but omit C

The definite integration results in a value.

Evaluating the Definite Integral

1

1

e.g.1

1

2

2 23 dxx


xx 23 e.g.1

1

2

2 23 dxx



1

2

Draw square brackets and hang the limits on the end




Replace x with

• the top limit

• the bottom limit

xx 23 e.g.1

1

2

2 23 dxx1

2



Subtract and evaluate

1 1 2 3I

32 2 2 2 12I ( ) ( )


xx 23 e.g.1

1

2

2 23 dxx1

2




So,

1

2

2 23 dxx

xx 23 e.g.1

1

2

2 23 dxx1

2

–12–3

= 15

Definite IntegrationSUMMAR

Y

Find the indefinite integral but omit C

Draw square brackets and hang the limits on the end

Replace x with

• the top limit• the bottom limit

Subtract and evaluate

The method for evaluating the definite integral is:

Definite IntegrationEvaluating the Definite

Integral

e.g. 2 Find

1

1

2 127 dxxx

Solution:

1

1

2 127 dxxx1

1

23

122

7

3

x

xx

Indefinite integral but no C

Definite IntegrationEvaluating the Definite

Integral

e.g. 2 Find

1

1

2 127 dxxx

Solution:

1

1

2 127 dxxx1

1

23

122

7

3

x

xx

1

1 712

3 2I

Substitute for x: top limit minus bottom

limit

3 2

1

1 7 112 1

3 2I

( ) ( )( )

51 68I 5

1 615I Simplify

1

1

2 127 dxxx 1 1 I I5 5 26 6 38 15 24

Definite IntegrationExercise

s

1. Find 2

1

2 143 dxxx

2. Find

2

2

2 326 dxxx

122141212)2(2)2( 23

)2(3)2()2(2)2(3)2(1)2(2 2323

64166416 20)14(6

2

1

23

2

4

3

3

x

xx 2

11

1

2

2

23

32

2

3

6

x

xx2 1

1 1

I2 I1

I2 I–2


26: Definite Integration and Areas

“Teach A Level Maths”

Vol. 1: AS Core Modules


0 1

23 2 xy

It can be used to find an area bounded, in part, by a curve

e.g.

1

0

2 23 dxx gives the area shaded on the graph

The limits of integration . . .

Definite integration results in a value.

Areas


. . . give the boundaries of the area.

The limits of integration . . .

0 1

23 2 xy

It can be used to find an area bounded, in part, by a curve

Definite integration results in a value.

Areas

x = 0 is the lower limit( the left hand

boundary )x = 1 is the upper limit(the right hand

boundary )

dxx 23 2

0

1

e.g.

gives the area shaded on the graph


0 1

23 2 xy

Finding an area

So the shaded area equals 3The units are usually unknown in this type of

question

1

0

2 23 dxxSince

1

0

xx 23

A1= 3 A0= 0

Area = A1–A0 = 3


Y

• the

curve

),(xfy

• the lines x = a and x = b

• the x-axis and

PROVIDED that the curve lies on, or above, the x-axis

between the values x = a and x = b

The definite integral or

gives the area between

b

a

dxxf )( b

a

dxy


xxy 22 xxy 22

Finding an area

0

1

2 2 dxxxA area

A B

For parts of the curve below the x-axis, the definite integral is negative, sofind area

12

0

B x 2x dx and ignore the sign


xxy 22

A

Finding an area

0

1

2 2 dxxxA

2

2

3

23 0

1

xx

32

0

10 1

3-1

( ) A ( )A

130 1Area

1

1

3

4Area A


xxy 22

B

Finding an area

12

0

2B x x dx

2

3 1

03

xx

1

11 0

3 0 AA

2

3Area

2

3 ignore the -ve signArea B


Y An area is always positive.

The definite integral is positive for areas above the x-axis but negative for areas below the axis.

To find an area, we need to know whether the curve crosses the x-axis between the boundaries.• For areas above the axis, the definite

integral gives the area.• For areas below the axis, we need to

change the sign of the definite integral to find the area.


Exercise Find the areas described in each

question.1. The area between the curve the x-

axis and the lines x = 1 and x = 3.

2xy

2. The area between the curve , the x-axis and the x = 2 and x = 3.

)3)(1( xxy


B

)3)(1( xxy

A

2xy

1.

2.

Solutions: 3

1

3

1

32

3

xdxxA

3

232

3

3

2

23

xx

x

3 2

2B x 4x 3dx

( )2

Area B ignore sign3

32

338

3

)1(

3

)3(


22 xxy

xy

Harder Arease.g.1 Find the coordinates of the points of

intersection of the curve and line shown. Find the area enclosed by the curve and line.

22 xxx

Solution: The points of intersection are given by

02 xx 0)1( xx10 xx or


22 xxy

xy 00 yx

xy Substitute in

11 yx The area required is the area under the curve between 0 and 1 . . . . . . minus the area under the line (a

triangle )

3

2

32

1

0

32

1

0

2

x

xdxxx

Area of the triangle 2

1)1)(1(

2

1

Area under the curve

Required area 6

1

2

1

3

2

Method 1

0 1


22 xxy

xy Instead of finding the 2 areas and then subtracting, we can subtract the functions before doing the integration.

x x dx 1

2

0

Area

Top curve – bottom curve =

Method 2

xxx 222xx

1 0

1 1Area A A 0

2 3

6

1

0 1

x x

12 3

02 3


6y

22 xy

Exercise Find the points of intersection of the

following curves and lines. Show the graphs in a sketch, shade the region bounded by the graphs and find its area.

22 xy 6y(a) ; (b) ; 2xy24 xy

Solution:

(a) 622 x

42 x2x

( y = 6 for both points )


6y

22 xy

Shaded area = area of rectangle – area under curve

Area A A2 2

Area under curve

2

2

32

2

2 23

2

x

xdxx

Shaded area = area under rectangle – area under

curve 1

324 13 2310

13

8 84 4 13

3 3


6y

22 xy

Area A A2 2

Area

x dx

2

2

2

4

23

8 88 8 10

3 3

Alternatively

Subtracting the functionsTop curve – bottom curve = 6 - (x2 + 2)

= 6 – x2 – 2

= 4 – x2

xx

23

2

43


2xy

24 xy

,02 yx

Area of the triangle

242 xx

2x 1xor

Substitute in : 2xy31 yx

Area under the curve

1

2

31

2

2

344

x

xdxx 9

3321

(b) ; 2xy24 xy

022 xx0)1)(2( xx

Shaded area = area under curve – area of triangle

29

29

At intersection


2xy

24 xy

Area 22 - x - xdx

1

2

Top curve – bottom curve = (4 – x2) - (x + 2)

Alternatively

Subtracting the functions

= 4 – x2 – x – 2

= 2 – x2 – x

A1=11/6

A–2= –31/3

Area A A 11 2 24

x xx

13 2

2

23 2


3xy

The symmetry of the curve means that the integral from 1 to +1 is 0.

If a curve crosses the x-axis between the limits of integration, part of the area will be above the axis and part below.

3xy e.g. between 1 and +1

To find the area, we could integrate from 0 to 1 and, because of the symmetry, double the answer.For a curve which wasn’t symmetrical, we

could find the 2 areas separately and then add.

Definite IntegrationYou don’t need to know how the formula for

area using integration was arrived at, but you do need to know the general ideas.

The area under the curve is split into strips.

The area of each strip is then approximated by 2 rectangles, one above and one below the curve as shown.

The exact area of the strip under the curve lies between the area of the 2 rectangles.


Using 10 rectangles below and 10 above to estimate an area below a curve, we have . . .Greater accuracy would be given with 20 rectangles below and above . . .For an exact answer we let the number of rectangles approach infinity. The exact area is “squashed”

between 2 values which approach each other. These values become the

definite integral.

25: definite integration © christine crisp “teach a level maths” vol. 1: as core modules

Documents

definite integral slide

definite integration

definite integration

definite integration

limits of integration

area area b slide

c slide

sign slide