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50: Harder 50: Harder Indefinite Indefinite Integration Integration © Christine Crisp Teach A Level Maths” Teach A Level Maths” Vol. 1: AS Core Vol. 1: AS Core Modules Modules

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

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Page 1: “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

Page 2: “Teach A Level Maths” Vol. 1: AS Core Modules

Indefinite Integration

Module C2

"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

Page 3: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 4: “Teach A Level Maths” Vol. 1: AS Core Modules

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.

Page 5: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 6: “Teach A Level Maths” Vol. 1: AS Core Modules

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:

Page 7: “Teach A Level Maths” Vol. 1: AS Core Modules

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 ).

Page 8: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 9: “Teach A Level Maths” Vol. 1: AS Core Modules

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.

Page 10: “Teach A Level Maths” Vol. 1: AS Core Modules

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.

Page 11: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 12: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 13: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 14: “Teach A Level Maths” Vol. 1: AS Core Modules

Indefinite Integration

Page 15: “Teach A Level Maths” Vol. 1: AS Core Modules

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.

Page 16: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 17: “Teach A Level Maths” Vol. 1: AS Core Modules

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:

Page 18: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 19: “Teach A Level Maths” Vol. 1: AS Core Modules

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

Page 20: “Teach A Level Maths” Vol. 1: AS Core Modules

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,

Page 21: “Teach A Level Maths” Vol. 1: AS Core Modules

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