higher higher unit 2 what is integration the process of integration integration & area of a...

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Higher

Higher Unit 2Higher Unit 2

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What is Integration

The Process of Integration

Integration & Area of a Curve

Exam Type Questions

Higher Outcome 2

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Higher

Integration

Part 1 Anti-Differentiation

Integration can be thought of as the opposite of differentiation

(just as subtraction is the opposite of addition).

In general: Differentiating Integrating

1( )n ndx nx

dx

1

1

nn xx dx C

n

Confusing? Is there any easier way?

Outcome 2

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Differentiation

multiply by power

decrease power by 1

Integration

increase power by 1

divide by new power

nx

1

1

nn xx dx

nC

Where does this + C come from?

IntegrationOutcome 2

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Integrating is the opposite of differentiating, so:

( )f x ( )f xintegrate

2

2

2

( ) 3 1

( ) 3 4

( ) 3 10

f x x

f x x

f x x

( )

( )

( )

f x

g x

h x

But: differentiate

differentiate

integrate

Integrating 6x….......which function do we get back to?

6x


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Solution:

When you integrate a function

remember to add the

Constant of Integration……………+ C


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Higher

6x dx means “integrate 6x with respect to x”

( )f x dx means “integrate f(x) with respect to x”

Notation

This notation was “invented” by

Gottfried Wilhelm von Leibniz


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Higher

Examples:

7x dx8

8x

23 2 1x x dx 3 2

3 23 2

x x

x C

3 2 x x x C


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Higher

2233

1 14 3x dxx

x

2 32 23 2(12 4 3 )

x x x x dx

1 53 13 2

513 2

12 4 33 1

x x x x

1 53 163 2

54 12 x x x x C


Just like differentiation, we must arrange the

function as a series of powers of x

before we integrate.

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Higher

2If '( ) 3 4 1 and (2) 11, find ( ).F x x x F F x To get the function F(x) from the derivative F’(x)

we do the opposite, i.e. we integrate.

2( ) (3 4 1)F x x x dx 3 2

3 43 2

x xCx

3 22x x Cx

3 2

(2) 11

2 2.2 2 11

C

F

8 8 2 11 C

3 C

3 2( ) 2 3 F x x x xHence:


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Further examples of integration

Exam Standard


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Higher

The integral of a function can be used to determine the area between the x-axis and the graph of the

function.

x

y

ba

( )y f x

Area ( )abf x dx

( ) ( ) ( ) ( ) ( )a

baf x f x dx F Fdx F bx If then

NB: this is a definite integral.

It has lower limit a and an upper limit b.

Area under a CurveOutcome 2

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Examples:

5

1(3 )x dx ( ) (3 )F x x dx

2

( ) 3 ( )2

xF x x c

5

1(3 ) (5) (1)x dx F F 25 1

15 32 2

30 25 6 1

248

2 2

2( 2)x dx

2( ) ( 2)F x x dx

3

( ) 2 ( )3

xF x x c

2 2

2( 2) (2) ( 2)x dx F F

8 84 4

3 3

168

3

8

3


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Higher

( )From the definition of it follows that:b

a

f x dx

( ) ( )b af x dx f x dx

a b

( ) ( ) ( ( ) ( ) )F b F a F a F b

Conventionally, the lower limit of a definite integral

is always less then its upper limit.


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a bc d

y=f(x)

( ) 0b

a

f x dx ( ) 0d

c

f x dx

Very Important Note:

When calculating integrals:

areas above the x-axis are positive areas below the x-axis are negative

When calculating the area between a curve and the x-axis:• make a sketch

• calculate areas above and below the x-axis separately

• ignore the negative signs and add


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a b

( )y f x

( )y g x

The Area Between Two Curves

To find the area between two curves we evaluate:

( )top curve bottom curveArea

( ( ) ( )b

a

Area f x g x dx


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Higher Example:

2 2

2, 4

4

Calculate the area enclosed by the lines

and the curves and

x x

y x y x

4 42 2 2

2 2

[ (4 )] (2 4)Area x x dx x dx 3

2 2(2 4) ( ) 4

3

xx dx F x x

(4) (2)Area F F

128 16( 16) ( 8)

3 3112

83

88

3

2y x

24y x

2x

4x

y

x


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Higher Complicated Example:

The cargo space of a small bulk carrier is 60m long. The shaded part of the diagram represents the uniform cross-section of this space.

2

, 6 64

1 9.

xy x

y y

It is shaped like a parabola with ,

between lines and

Find the area of this cross-section and hence find the volume of cargo that this ship can carry.


9

1

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1y

9y

2

4

xy

x

y

ss

The shape is symmetrical about the y-axis. So we calculate the area of one of the light shaded rectangles and one of the dark shaded wings. The area is then

double their sum.t t

The rectangle: let its width be s2

14

2ss

y

The wing: extends from x = s to x = t

2

694

ty t

The area of a wing (W ) is given by:

2

(9 )4

t

s

xW dx

Area under a Curve

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6 2

2

(9 )4

xW dx

2 3

(9 ) ( ) (9 )4 12

x xdx F x x

6.6.6(6) (9.6 ) 54 18 36

12F 2.2.2 2

(2) (9.2 ) 1812 3

F

2 56(6) (2) 36 18

3 3W F F

The area of a rectangle is given by:

(9 1) 16R s

The area of the complete shaded area is given by:

2( )A R W

56 48 56 2082(16 ) 2

3 3 3A

The cargo volume is: 320860 60. 20.208 41

360A m


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Higher

Exam Type QuestionsOutcome 2

At this stage in the course we can only do

Polynomial integration questions.

In Unit 3 we will tackle trigonometry integration

Integration

Higher Mathematics

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222

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34

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91

higher higher unit 2 what is integration the process of integration integration & area of a...

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