AS Maths Masterclass

Lesson 4:

Areas by integration and the Trapezium rule

Learning objectives

The student should be able to:

• understand why integration is concerned with finding areas under curves;

• apply the integration rule to simple polynomials to find areas under curves;

• evaluate approximate areas using the trapezium rule;

Revision of C1 integration

Can you still remember how to integrate?

Recall that

where c is a constant of integration.

You should also remember that integration is the reverse process of differentiation.

Click here for some revision questions: integration to find the constant.

cn

xdxx

n

n

1

1

A first principles approach to area.

Consider finding the area under the graph

f (x) = 2x + 3, between the lines x = 0 and x = 10.

We can divide the area into

n strips, and so the width

of each strip will be

Now, and so on

For the last strip we have

n

103

23

10

f (x) = 2x + 3

3)0( f 3)10(2)

10(

nnf 3)

20(2)

20(

nnf

3)10

10(2)10

10( nn

f

Finding the sum

If we now calculate the sum of these n terms we get:

And so:

Did you notice the AP ?

a = 3,

)10.(3

2020....)

10).(3

40()

10).(3

20()

10(3

nnnnnnnSn

)3

)1(20(....)3

40()3

20(3

10

n

n

nnnSn

nd

20

Using the AP formula

If we now use the AP formula: a=3,

We have

Hence,

So

nd

20

dnan

Sn

)1(22

nn

n

nSn

20).1(6

2.

10

)1(

2065 n

nSn

)1(100

30 nn

n

10010030

n

100130

nas130

Validation of the area

We conclude that the area is 130 square units.

Of course, this area could have been calculated from a single trapezium, but when the function f (x) is curved, a trapezium would only provide an approximation to the precise area.

Extension activity: Now take and perform

a first principles approach for the area under the curve between x = 1 and x = 5.

2)( xxf

Connecting up the resultsIn the first case we saw that when f (x) = 2x+3 we obtained an area of 130 square units.

But

If we defineThen F ( 10 ) = 130, F ( 0 ) = 0 and F ( 10 ) – F ( 0 ) = 130

In the second case we saw that when we obtained an area of square units.

And again, defining then F ( 5 ) = and F ( 1 ) =

So, as before F ( 5 ) – F ( 1 ) =

Cxxdxx 3)32( 2

xxxF 3)( 2

3)(

3xxF

3141

2)( xxf

3241 3

1

3141

Definite integration practice

The three links below will help you practice evaluating a range of definite integrals:

• Click here for the integration of quadratics

• Click here for the integration of polynomials

• Click here for ones with negative indices and surds

Turning our attention to area …

Click on this link to see a range of integration approaches in practice

We shall talk about the trapezium rule later, but for now let’s investigate finding the area under a given curve y = f ( x ) between x = a and x = b.

First principles in general form:If we take a small rectangular strip MNQP where P is the point (x, y), and N has x co-ordinate for small , then as gets smaller, the rectangle will approximate to the area under the curve and between

x = M and x = N.

The small section of the area

can be called

Now, if we divide up the whole required area that falls between x = a and x = b into many strips, then for each strip:

xx x x

Aa bM N

P Q

y

x

A xy

Integration connections

Hence, total area = for small

So taking even smaller, Area =

But earlier we agreed that so

Now, taking the limit once again,

But by definition, and so

Hence, because we know from previous work that integration and differentiation are reverse processes.

b

ax

xyA xx

b

axxxy

lim

0

A xy yx

A

yx

Ax

lim

0

dx

dA

x

Ax

lim

0y

dx

dA

dxyA

The area function and finding areas between line and curve.

Click here to see how an Area function can be obtained

Click here to practice finding the area between oblique lines and curves

Click here for more practice

Towards the Trapezium rule

As a crude first approximation to the areabounded by the lines x = a, x = b and thecurve y = f (x), we could take an estimate:

either the rectangle ABCD or the rectangleABEF. With a little thought, however, boththese approximations can be improvedupon by taking the trapezium ABED.

Hence, areaClearly, the smaller the width of the interval, the better theapproximation to the precise area. We can therefore take advantage

ofthis fact and divide the interval into two halves. If we let m denote themidpoint of the interval (a,b) ….

)()(2

bfafab

Finding the formula

Then area

where h = m – a = b – m

And by repeated subdivision into n trapezia (activity ?) we can easily obtain

Area

)()(2

)()(2

bfmfmb

mfafam

)()()()(2

bfmfmfafh

)()(2)(2

bfmfafh

)}()(2....)2(2)(2)({2

bfhbfhafhafafh

The trapezium rule

This rule is easier to understand if we change the notation

as shown in the diagram

This is what we mean when we talk

about the trapezium rule.

Area

where

nnn yyyyyyh

12210 22...222

n

abh

How it works in practice

Click here to see the trapezium rule in

practice

Click here to see a comparison of numerical

methods

An example

Let’s evaluate using the trapezium rule with

n = 4 divisions.

Integral

1.4463

This estimate is an over-estimate of the precise area

1.442695… in this particular case. (why ?)

dxx1

0

2

hb a

n

1 0

40 25.

0 125 0 2 0 25 2 0 5 2 0 75 1. [ ( ) ( . ) ( . ) ( . ) ( )]f f f f f

]23636.38284.23784.20.1[125.0

Tip of the day

It is sensible to tabulate the working in order that any human

errors in the calculation can be minimised.

e.g. Using six equally spaced ordinates with the trapezium

rule, find an estimate for

Now, 6 ordinates means 5 trapezia so n = 5, a = 1, b = 6

and

We let and form the table:

dxx6

1

2110 )(log

15

16

n

abh

2/1

10)(log xy

Table method

i xi yi 2yi

0 1 =0

1 2 =0.54866 1.09732

2 3 =0.69074 1.38148

3 4 =0.77593 1.55185

4 5 =0.83604 1.67209

5 6 =0.88213

0.88213 5.70274

2110 )1(log

2110 )2(log

2110 )3(log

2110 )4(log

2110 )5(log

2110 )6(log

Final calculation

Hence,

Area (2

1 0.88213 + 5.70274 ) = 3.2924

Click here to practice questions on the trapezium rule

Click here for more practice