Top Banner
AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule
21

AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

Jan 12, 2016

Download

Documents

Jeffry McKinney
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

AS Maths Masterclass

Lesson 4:

Areas by integration and the Trapezium rule

Page 2: 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;

Page 3: AS Maths Masterclass Lesson 4: Areas by integration and 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

Page 4: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 5: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 6: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 7: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 8: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 9: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 10: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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.

Page 11: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 12: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 13: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 14: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 15: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 16: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 17: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

How it works in practice

Click here to see the trapezium rule in

practice

Click here to see a comparison of numerical

methods

Page 18: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 19: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 20: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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

Page 21: AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule.

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