introduction integration is the reverse process of differentiation differentiating gives us a...

Introduction

• Integration is the reverse process of Differentiation

• Differentiating gives us a formula for the gradient

• Integrating can get us the formula for the curve, if we know the gradient function

• It can also be used to calculate the Area under a curve

IntegrationYou can integrate functions of the form f(x) = axn where ‘n’ is

real and ‘a’ is a constant

Integrating is the reverse process of differentiation. Let us think about a differentiation for a moment.

If:

If:

If:

8A

2y x

2dy

xdx

2 5y x

2dy

xdx

2 - 7y x

2dy

xdx

So integrating 2x should give us x2, but we will be unsure as to whether a number has been added or

taken away

Differentiating

Function

Multiply by the power

Reduce the power by 1

Gradient Function

Integrating

Function

Divide by the power

Increase the power by 1

Gradient Function




8A

Integrating

Function

Divide by the power


Gradient Function

Mathematically speaking…

ndyx

dx

1

1

nxy

n

We increased the power by 1, then divided by the

(new) power

If:

Then:




8A

Integrating

Function

Divide by the power


Gradient Function

Example Questions

ndyx

dx

1

1

nxy

n

If:

Then:

Integrate the following:

a)4dyx

dx

5

5

xy c

Increase the power by one, and divide by the new power

DO NOT FORGET TO ADD C!

b)5dy

xdx

4

4

xy

c



41

4y x c




8A

Integrating

Function

Divide by the power


Gradient Function

Example Questions

ndyx

dx

1

1

nxy

n

If:

Then:

Integrate the following:

c)32

dyx

dx

42

4

xy c



d)1

23dy

xdx

3

233

2

xy c



4

2

xy c

3

22y x c

IntegrationYou can apply the idea of

Integration separately to each term of dy/dx

In short, if you have multiple terms to integrate, do them all separately

8B

Example QuestionIntegrate the following:

13 26 2 3

dyx x x

dx

26

2

xy

22

2

x

3

23

32

x

23y x 2 x3

2 2x c

Integrate each part separately

‘Tidy up’ terms if possible

IntegrationYou can apply the idea of

Integration separately to each term of dy/dx

In short, if you have multiple terms to integrate, do them all separately

8B

Example QuestionIntegrate the following:

1 3

2 21 1

'( )2 2

f x x x

1

2121

2

xy

1

212

12

x


1

2y x1

2 x

y x1

x

Deal with the

fractions

Rewrite if necessary

c

c

IntegrationYou need to be able to use

the correct notation for Integration

8C

nx dx

This the the integral sign,

meaning integrate

This is the expression to be

integrated (brackets are often used to separate

it)

The dx is telling you to integrate ‘with respect to

x’

Example QuestionFind:

132( 2 ) x x dx

3

2

3

2

x

42

4

x

3

22

3x 41

2x c


Deal with the

fractions



8C

nx dx


meaning integrate



it)

The dx is telling you to integrate the

‘x’ parts


3

2( 2) x dx

1

2

1

2

x

2x

1

2 2x

2x c


Deal with the

fractions



8C

nx dx


meaning integrate



it)


‘x’ parts


2 2 2(3 ) x px q dx

33

3

x

1

1

px


2 q x

3 x 1 px 2 q x c

Deal with the

fractions

p and q2 should be treated as if they were just numbers!



8C

nx dx


meaning integrate



it)


‘x’ parts


2(4 6) t dt

34

3

t 6t


c

IntegrationYou can find the constant of

integration, c, if you are given a point that the function passes

through

Up until now we have written ‘c’ when Integrating.

The point of this was that if we differentiate a number on its own, it disappears.

Consequently, when integrating, we cannot be sure whether a number was there originally, and what it was if there was one…

Step 1: Integrate as before, putting in ‘c’Step 2: Substitute the coordinate in, and work out what ‘c’ must be to make the equation balance.. 8E

Example QuestionThe curve X with equation y = f(x)

passes through the point (2,15). Given that:

Find the equation of X.

2'( ) 5 3f x x x

2'( ) 5 3f x x x Integrate

3 25 3( )

3 2

x xf x c

Sub in (2,15)3 25(2) 3(2)

153 2

c

40 1215

3 2c

115 19

3c

1 4

3c

3 25 3 1( ) 4

3 2 3

x xf x

Work out each fraction

Add the fractions together

Work out c



through








2 2'( )

xf x

x

2 2'( )

xf x

x

2

'( )x

f xx

2

x

3

2'( )f x x1

2 2x

5

2

( )5

2

xf x

1

22

12

x

5

22

( )5

f x x1

2 4x + c

Split into 2 parts

Write in the form axn

Integrate



through








2 2'( )

xf x

x

5

22

( )5

f x x1

2 4x + c

52( )

5f x x 4 x + c

525 4

5 4 4 + c

5 12.8 8 + c

0.2 c

Rewrite for substitution

y = 5, x = 4

Work out each part carefully

5

22

( )5

f x x1

2 4x + 0.2

Summary

• We have learnt what Integration is

• We have seen it combined with rewriting for substitution

• We have learnt how to calculate the missing value ‘c’, and why it exists in the first place

introduction integration is the reverse process of differentiation differentiating gives us a...

Documents

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