Higher Maths 2 2 Integration

1

Speed Time Graphs

D

S T×÷÷

20

40

00 2 4

Time (hours)

Sp

eed

(mp

h)

Calculate the distance travelled in each journey.

20

40

00 2 4 6Time (hours)

Sp

eed

(mp

h)

20

40

00 2 4 6Time (hours)

Sp

eed

(mp

h)

D=4 × 30= 120

miles

average speed

D=5 × 20= 100

miles

30 miles 90 miles + 15

miles 135 miles

In Speed Time graphs, the distance travelled is the same as the area under the graph.

Higher Maths 2 2 Integration

2

Reverse Differentiation

D

Tspeed

=

‘rate of change ofdistance with respect to time’

If we know how the speed changes, and want to find distance,we need to ‘undo’ finding the rate of change with respect to time.In other words we need to reverse differentiate.Differentiating backwards is used to find the area under a function.

f (x)

Higher Maths 2 2 Integration

3

Estimating Area Under Curves

To estimate area under a function, split the area into vertical strips.

f (x)

x

f (x)

x

The area of each strip is the height, multiplied by :x

Total Area

f (x) x×

f (x) x×( )=

As the strips get narrower, the estimate becomes more accurate.

Area under the function=

f (x)

f (x) x as 0

means ‘the sum of...’

Higher Maths 2 2 Integration

4

x

Higher Maths 2 2 Integration

5

IntegrationThe algebraic method for finding area under a function is called Integration.

‘Integrate’means ‘join together all the pieces’

Integration uses reverse differentiation to ‘undo’ finding the rate of change.

∫Area under the function

=

f (x)

f (x) x

d x

xas 0

f (x)=

x

∫ d x

Expression or function to be integrated.

Higher Maths 2 2 Integration

6

Differentiating BackwardsIntegration involves differentiating in reverse.

multiply by the power

reduce the power by one

f (x) f ′(x)

divide by the power

increase the power by one

f (x) f ′(x)

• divide every x-term by the new power

How to differentiate backwards:• increase the power of every x-term by one

Higher Maths 2 2 Integration

7

Finding the Anti-DerivativeThe result of a

differentiation is called a

derivative.

• divide by each new power

How to Reverse Differentiate• increase each power by one

The result of differentiating

backwards is called the anti-derivative.

Example

dydx

= 8 x 3 + x

2 – 6

x

Find the anti-derivative

for

y

y = ∫ dx8 x 3 + x

2 – 6 x

=2 x

4 + x 3

– 3

x2

13

∫ d x expression or function

Higher Maths 2 2 Integration

8

The Constant of IntegrationWhen differentiating, part of an expression is

often lost.Example

f (x)= x 4 + 2

f (x) = x 4

f (x) = x 4 – 7

f ′(x) = 4 x 3

All three functions have derivative

The anti-derivative of

f ′(x) is

f (x)= x 4 + c

unknown

constant

When differentiating in

reverse,

it is essential to remember

to add back on the unknown

number.This is called the constant of

integration.

Higher Maths 2 2 Integration

9

Basic IntegrationThe result of

integration is called

an integral.

f (x)

Example 1

∫ d xf (x)= x 6

• divide by each new power

How to Integrate

• increase each power by one

=

17= x

7 + c

∫ d xx 6

• add the constant of integration

constant of integration

Example 2

∫ d x4x

Fin

d

= d x4 x12∫

= 8 x +

c

= 8 x12 + c

Higher Maths 2 2 Integration

10

Integration and AreaIntegration can be used to find the area ‘under’ a

function between two different values of x.

f (x)

x1 x2

∫ d xf (x)

x1

x2

= ∫ d xf ( x2) – ∫ d xf ( x1)

Area under f (x) between x1

and x2

=

‘Upper Limit’

‘Lower Limit’

This is called a Definite Integral

The area ‘under’ a function can be described more mathematically as the area between the function and the

x-axis.

Area Between a Function and the x-axis

Higher Maths 2 2 Integration

11

f (x)

∫ d xf (x)

5

-2

5-2

g (x)

∫ d xg(x)

3

-6

3-6

h (x)

8-7

∫ d xh(x)

8

-7

Higher Maths 2 2 Integration

12

Evaluating Definite Integrals

=∫ dx2 x 3

0

3

[ x 4 + c1

2 ]0

3

= ( × ( 3 ) 4 + c1

2 ) ( × ( 0 ) 4 + c1

2 )–

= 4012

The constants of integration cancel each other out.

Definite Integrals do not require the constant of integration.

∫ d xx1

x2

Definite Integral

Write integral inside square

brackets

units 2

Example

[

]

Higher Maths 2 2 Integration

13

Evaluating Definite Integrals (continued)

Example 2

4

1∫ d x

4

19 x

2 – 2 x 3 x 3 – x

2=

( )3 × 43 –

42

= – ( )3 × 13 –

12

( )192 – 16

= – ( )3 – 1

= 174

units

2

Find the area below the curve

between x = 1 and x = 4.

y = 9 x 2 – 2 x

Write integral inside square

brackets...

(no constant required)

...then evaluate for

each limit and subtract.

Remember units!

When calculating areas by integration, areas above

the x-axis are positive and areas below the x-axis are negative.

Areas Above and Below the x-axis

Higher Maths 2 2 Integration

14

bac d

∫ dxf (x)

a

b

> 0

∫ dxf (x)

c

d

< 0

f (x)

How to calculate area

between

a curve and the

• draw a sketch

• calculate the areas above

and

below the axis

separately

• add the positive value of

each

area (ignore negative

signs)

x-axis :

x-

Area Between FunctionsIntegration can also be used to find the area between

two graphs, by subtracting integrals.

Higher Maths 2 2 Integration

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f (x)

g (x)

Area enclosed by f (x) and g

(x)

∫ d xf (x)

=

= – ∫ d xg (x)

∫ f (x) d xg (x)( – )

ba

b

a

b

a

b

a

between intersection points a

and b

is above

f (x) g (x)