The following slides show one of the 51 presentations that cover the AS Mathematics core modules C1 and C2. Demo Disc Teach A Level Maths Vol. 1: AS Core.

The following slides show one of the 51 presentations that cover the AS Mathematics core modules C1

and C2.

Demo DiscDemo Disc

““Teach A Level Maths”Teach A Level Maths”Vol. 1: AS Core ModulesVol. 1: AS Core Modules

AS MathematicsAS Mathematics

26: Definite 26: Definite Integration and AreasIntegration and Areas

© Christine Crisp

Definite Integration and Areas

Module C1

AQA Edexcel

OCR

MEI/OCR

Module C2

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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.


0 1

23 2 xy



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

the shaded area equals 3

The units are usually unknown in this type of question

1

0

2 23 dxxSince

31

0

xx 23


SUMMARY

• 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

1

0

2 2 dxxxB area

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


xxy 22

A

Finding an area

0

1

2 2 dxxxA

2

2

3

23 0

1

xx

2

3

)1(3

)1(0

1

3

1

1

1

3

4Area A


xxy 22

B

Finding an area

1

0

2 2 dxxxB

2

3 1

03

xx

013

1

3

2

3

2Area B

Definite Integration and Areas SUMMAR

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


A

2xy

1.

B

)3)(1( xxy

2.

Solutions:

3

1

3

1

32

3

xdxxA

3

232

3

3

2

23

xx

x

3

2

2 34 dxxxB

3

2 B

32

33

83

)1(

3

)3(


Extension

The area bounded by a curve, the y-axis and the lines y = c and y = d is found by switching the xs and ys in the formula.

So,become

sb

a

dxy

d

c

dyx

d

c

dyx

e.g. To find the area between the curve , the y-axis and the lines y = 1 and y = 2, we need

xy

3

7

2

1

2 dyy


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.

1

0

321

0

2

32

xx

dxxxArea

We get

Method 2

xxx 222xx

0

3

1

2

16

1

0 1


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

3

164

3

84

3

8

Area under curve

2

2

32

2

2 23

2

x

xdxx

Shaded area3

1624

3218


2xy

24 xy

,02 yx

Area of the triangle

2x 1xor

Substitute in : 2xy31 yx


1

2

31

2

2

344

xxdxx 9

3321

(b) ; 2xy24 xy

022 xx0)1)(2( xx

Shaded area = area under curve – area of triangle

29

29

242 xx


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 Integration and AreasYou 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.



The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.


23 2 xy

. . . give the boundaries of the area.



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

Definite Integration and AreasSUMMAR

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

1

0

2 2 dxxxB area

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

Definite Integration and Areas SUMMAR

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.

Definite Integration and AreasHarder

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


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


Required area 6

1

2

1

3

2

The following slides show one of the 51 presentations that cover the AS Mathematics core modules C1 and C2. Demo Disc Teach A Level Maths Vol. 1: AS Core.

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