Core 2 indefinite integration

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

1;1

1

nCn

xdxx

nn

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


e.g.1 Evaluate dxx

4

1

4

1

x

Solution: Using the law of indices,

4x

So, dxxdxx

44

1

Cx

3

3

Cx

3

3

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


e.g.1 Evaluate dxx

4

1

4

1

x


4x

So, dxxdxx

44

1

Cx

3

3

Cx

3

3

Cx

33

1But this one . . . is an index


e.g.2 Evaluate

Cx

23

23

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

2 gives

Cx

3

2 23

Cx

2

2

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:


Cx

23

e.g.2 Evaluate

Cx

23


2 gives

Cx

2

2

23

23

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

dxx 21

dxx 21

Solution: 23

2

3

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


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


e.g.4 Evaluate

dxx

x 1

Solution: dx

x

x 1dx

x

x

21

1

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

xx

x

x

We cannot integrate with x in the denominator.


e.g.4 Evaluate

dxx

x 1

Solution: dx

x

x 1dx

x

x

21

1

dxx 21

Cx 21

2

dx

xx

x

21

21

1

Instead of dividing by ,multiply by 2

332

3

2 23

x

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.


Solution:

dxx

xy 2

2 1

e.g.5 The curve passes through the point

( 1, 0 ) and

)(xfy

22/ 1

)(x

xxf Find the equation of the

curve.

( 1, 0 ) on the curve: C3

2

dxxxy 22

Cxx

y

13

13

Cx

xy

1

3

3

C 13

10

So the curve is 3

21

3

3

x

xy

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


Evaluate

dxxx )1(

Exercise

dxx 3

1

Solution:

dxxdxx 3

3

1

Cx

22

1

1. 2.

Cx

2

2

dxxxdxxx )1()1( 21

dxxx 2

123

Cxx

3

2

5

2 23

25


3. Given that , find the equation of

the curve through the point ( 2, 0 ).

2

2 1

x

x

dx

dy

Solution:

2

2 1

x

x

dx

dy dxxy 21

Cx

xy

1

1

Cx

xy 1


120 C

2

3

So the curve is 2

31

xxy

Exercise


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.


e.g.1 Evaluate dxx

4

1

4

1

x


4x

So, dxxdxx

44

1

Cx

3

3

Cx

3

3

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

Cx

33

1

But this one is an index


e.g.2 Evaluate

Cx

23

23


2 gives

Cx

3

2 23

Cx

2

2

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:


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:

dx

x

x

21

1

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

xx

x

x

21

21

1 x

x


dxx 21

Cx 21

23

2 23

x

21

x

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

dx

xx

x

21

21

1 So,


Solution:

dxx

xy 2

2 1

e.g.5 The curve passes through the point

( 1, 0 ) and .

)(xfy

22/ 1

)(x

xxf Find the equation of the

curve.


2

dxxxy 22

Cxx

y

13

13

Cx

xy

1

3

3

C 13

10

So the curve is 3

21

3

3

x

xy

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

Core 2 indefinite integration

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