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

24: Indefinite 24: Indefinite IntegrationIntegration

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

Indefinite Integration

Equal !

(a) 32xy xdx

dy2

xdx

dy2(b) 12xy

e.g.1 Differentiate (a) (b) 12 xy32 xy

The gradient functions are the same since the

graph of is a just a translation of 12 xy 32 xy

We first need to consider an example of

differentiation


12 xy

32 xy

At each value of x, the gradients of the 2 graphs are the same

xdx

dy2

e.g. the gradient at x = 1 is 2

Graphs of the functions


C is called the arbitrary constant

or constant of

integration

If we are given the gradient function and want to find the equation of the curve, we reverse the process of differentiation

xdx

dy2 Cxy 2So,

The equation forms a family of

curvesCxy 2

Indefinite integration is the reverse of differentiation

BUT the constant is unknown


• Make the power 1 more

To reverse the rule of differentiation:

e.g.2 Find the equation of the family of curves

which have a gradient function given by26xdx

dy

• Drop it thro` the trap door

36xy

Solution:

26xdx

dy


• Make the power 1 more

3

6 3xy

To reverse the rule of differentiation:

e.g.2 Find the equation of the family of curves

which have a gradient function given by26xdx

dy

• Drop it thro` the trap door

• add C

Tip: Check the answer by differentiating

1

2

Cxy 32

C

Solution:

26xdx

dy

Make the power 1 more and drop it through the trap door


32xy

52 3 xyy

dx

dy26x

dx

dy

26xdx

dy

Cxy 32

( Sample of 6 values of

C )

The gradient function

The graphs look like this:


Cxx

y 2

3 2

Solution: The index of x in the term 3x is 1, so adding 1 to the index gives 2.

13 xdx

dy

e.g. 3 Find the equation of the family of curves with gradient function

The constant 1 has no x. It integrates to x.

We can only find the value of C if we have some additional information


xxdx

dy43 2 1.

Find the equations of the family of curves with the following gradient functions:

Exercises

12

12 xxdx

dy2.

)3)(2( xxdx

dy3.

N.B. Multiply out the brackets first

Cxx

y 2

4

3

3 23

:Ans Cxxy 23 2

Cxxx

y 43

23

:Ans

2

11

1


xxdx

dy43 2 1.

Find the equations of the family of curves with the following gradient functions:

Exercises

12

12 xxdx

dy2.

)3)(2( xxdx

dy3.

Cxxy 23 2

Cxxx

y 43

23

:Ans

Cxxx

y 623

23

:Ans

Cxx

y 2

4

3

3 23

:Ans2

11

1

62 xxdx

dy


• Make the power 1 more• Drop it thro` the trap door• 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

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


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

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

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.


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

Documents

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