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

50: Harder Indefinite 50: Harder Indefinite IntegrationIntegration

© Christine Crisp

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

ModulesModules

Indefinite Integration

Module C2

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Indefinite Integration

• add 1 to the power• divide by the new power• add C

1;1

1

nCnxdxxn

n

Reminder:

n does not need to be an integer BUT notice that the rule is for nx

It cannot be used directly for terms such as nx

1

Indefinite Integration

e.g.1 Evaluate dxx 41

4

1x

Solution: Using the law of indices,

4x

So, dxxdxx

44

1

Cx

3

3

Cx

3

3

This minus sign . . . . . . makes the term negative.

Indefinite Integration

e.g.1 Evaluate dxx 41

4

1x

Solution: Using the law of indices,

4x

So, dxxdxx

44

1

Cx

3

3

Cx

3

3

Cx

33

1But this one . . . is an index

Indefinite Integration

e.g.2 Evaluate

Cx

23

23

We need to simplify this “piled up” fraction.Multiplying the numerator and denominator by

2 givesCx

3

2 23

Cx

22

23

23

We can get this answer directly by noticing that . . . . . . dividing by a fraction is the same as multiplying by its reciprocal. ( We “flip” the fraction over ).

dxx 21

dxx 21

Solution:

Indefinite Integration

Cx

23

e.g.2 Evaluate

Cx

23

We need to simplify this “piled up” fraction.Multiplying the numerator and denominator by

2 givesCx

22

23

23

We can get this answer directly by noticing that . . .

dxx 21

dxx 21

Solution: 23

23

. . . dividing by a fraction is the same as multiplying by its reciprocal. ( We “flip” the fraction over ).

Indefinite Integration

e.g.3 Evaluate dxx

1

xSolution:

21

x

So, dxx

dxx 2

1

11

Using the law of indices,

dxx 21

Cx

21

21

Cx 21

2

Indefinite Integration

e.g.4 Evaluate

dxx

x 1

Solution: dx

x

x 1 dxx

x

211

dxx 21

dx

xx

x

21

21

1

Write in index form xSplit up the fraction

Use the 2nd law of indices:

21

211

21

xxx

x

We cannot integrate with x in the denominator.

Indefinite Integration

e.g.4 Evaluate

dxx

x 1

Solution: dx

x

x 1 dxx

x

211

dxx 21

Cx 21

2

dx

xx

x

21

21

1

Instead of dividing by ,multiply by 2

332

3

2 23x

Instead of dividing by ,multiply by 22

1

21

x

and 21

210

21

1 xx

x

The terms are now in the form where we can use our rule of integration.

Indefinite Integration

Solution: dx

xxy

22 1

e.g.5 The curve passes through the point

( 1, 0 ) and

)(xfy

22/ 1)(

xxxf

Find the equation of the curve.

( 1, 0 ) on the curve: C32

dxxxy 22

Cxxy

13

13 C

xxy

13

3

C 1310

So the curve is 3

213

3

xxy

It’s important to prepare all the terms before integrating any of them

Indefinite Integration

Evaluate

dxxx )1(

Exercise

dxx 31

Solution:

dxxdxx 3

31

Cx

22

1

1. 2.

Cx

2

2

dxxxdxxx )1()1( 21

dxxx 2

123

Cxx

32

52 2

325

Indefinite Integration

3. Given that , find the equation of

the curve through the point ( 1, 0 ).

2

2 1xx

dxdy

Solution: 2

2 1xx

dxdy

dxxy 21

Cxxy

1

1C

xxy

1

( 2, 0 ) on the curve: C2120 C

23

So the curve is 2

31

xxy

Exercise

Indefinite Integration

Indefinite Integration

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.

Indefinite Integration

e.g.1 Evaluate dxx 41

4

1x

Solution: Using the law of indices,

4x

So, dxxdxx

44

1

Cx

3

3

Cx

3

3

This minus sign . . . . . . makes the term negative.

Cx

33

1But this one is an index

Indefinite Integration

e.g.2 Evaluate

Cx

23

23

We need to simplify this “piled up” fraction.Multiplying the numerator and denominator by

2 givesCx

3

2 23

Cx

22

23

23

We can get this answer directly by noticing that . . . . . . dividing by a fraction is the same as multiplying by its reciprocal. ( We “flip” the fraction over ).

dxx 21

dxx 21

Solution:

Indefinite Integration

e.g.3 Evaluate dxx

1

xSolution:

21

x

So, dxx

dxx 2

1

11

Using the law of indices,

dxx 21

Cx

21

21

Cx 21

2

Indefinite Integratione.g.4

Evaluatedx

x

x 1

Solution:

dxx

x

211

dx

xx

x

21

21

1

Write in index form x

Split up the fraction

We cannot integrate with x in the denominator.

Use the laws of indices: and21

211

21

xxx

x

21

21

1 x

x

Indefinite Integration

dxx 21

Cx 21

23

2 23x

21

x

The terms are now in the form where we can use our rule of integration.

dx

xx

x

21

21

1 So,

Indefinite Integration

Solution: dx

xxy

22 1

e.g.5 The curve passes through the point

( 1, 0 ) and .

)(xfy

22/ 1)(

xxxf

Find the equation of the curve.

( 1, 0 ) on the curve: C32

dxxxy 22

Cxxy

13

13 C

xxy

13

3

C 1310

So the curve is 3

213

3

xxy

It’s important to prepare all the terms before integrating any of them