higher higher unit 2 what is integration the process of integration ( type 1 ) area between to...

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Higher

Higher Unit 2Higher Unit 2

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What is Integration

The Process of Integration ( Type 1 )

Area between to curves ( Type 4 )

Outcome 2

Area under a curve ( Type 2 )

Working backwards to find function ( Type 5 )

Area under a curve above and below x-axis ( Type 3)

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Higher Outcome 2

Integration

1

( 1)

nn xx dx c

n

43x dx 53

5

xc

4 13

(4 1)

xc

we get

You have 1 minute to come up with the rule.

43x dx 53

5

xc

Integration can be thought of as the opposite of differentiation

(just as subtraction is the opposite of addition).

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Higher

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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Higher

Integrating is the opposite of differentiating, so:

( )f x ( )f xintegrate

2

2

2

( ) 3 1

g( ) 3 4

h( ) 3 10

f x x

x x

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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Higher

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

5 3

1 4

2x dx

x x

1 352 2( 4 )

2

xx x dx

3 14 2 2

3 12 2

48

x x xC


Just like differentiation, we must arrange the

function as a series of powers of x

before we integrate.

3

2

142

1 2 8

8 3

xC

xx

3

4

1 2 8

8 3

xC

x x

Name :

34x dx 2

2

3dx

x3 x dx 3 2

1

2dx

x 3

2

3

xdx

x

( 2)(3 1) x x dx 2 3

x

dxx

41

5 x dxx

Integration techniquesArea

under curve=

Area under curve=

IntegrationIntegration

44

4

xc

1

23x dx3

2

3

2

3xc

3

22x c

22

3

xdx

12

3( 1)

xc

2

3c

x

2

3

2

xdx

1

33

2

xc

323

xx dx

2 22

6 2

x xc

2

2

1

6

xc

x

1 1

2 45 x x dx

1 5

2 4

1 5

2 4

5x xc

1 5

2 42 4x x c

23 5 2 x x dx 3 23 5

23 2

x xx c

1 1

2 22 3x x dx

4x c

23 5

22

xx x c

3 1

2 2

3 1

2 2

2 3x xc

3122

46

3

xx c

Real Application of Integration

Find area between the function and the x-axis

between x = 0 and x = 5

A = ½ bh = ½x5x5 = 12.5

5

0

A x dx52

02

x

2 25 0 250 12.5

2 2 2




A = ½ bh = ½x4x4 = 8

4

0

4 A x dx 42

0

42

xx

24

4 4 0 8 16 242

A = lb = 4 x 4 = 16

AT = 8 + 16 = 24




22

0

A x dx23

03

x

3 32 0 8 80

3 3 3 3



between x = -3 and x = 3

3

3

A x dx

32

32

x

2 23 ( 3) 9 90

2 2 2 2

?

Houston we have a problem !

We need to do separate integrations for above and below the x-axis.


3

0

A x dx32

02

x

23 9 9

0 02 2 2

0

3

A x dx

02

32

x

2( 3) 9

02 2

Areas under the x-axis ALWAYS give negative values

By convention we simply take the positive value since we cannot get a negative area.

9 99

2 2TA

Integrate the function g(x) = x(x - 4) between x = 0 to x = 5


We need to sketch the function and find the roots before we

can integrate

We need to do separate integrations for above and below the x-axis.


4

0

( 4) A x x dx 43

2

0

23

xx

3

24 64 322 4 0 32

3 3 3

5

4

( 4) A x x dx 3 3

2 25 4 25 32 2 5 2 4

3 3 3 3

Since under x-axis

take positive value

32

3

7

3


4 5

0 4

( 4) ( 4) TA x x dx x x dx 32 7 13

3 3

Find upper and lower limits.

Area between Two Functions

12

0

( ) A x x dx 12 3

02 3

x x

2 31 1 1

0 2 3 6

1 6

2x x2 0x x

( 1) 0x x 0 1x x

then integrate

top curve – bottom curve.

y x2y x

Find upper and lower limits.


12 2

1

( 1) ( -1) A x x dx

2 21 1x x 22 2 0x

2( 1)( 1) 0x x 1 1x x then integrate

top curve – bottom curve.

22( 1) 0x

12

1

( 2 2) x dx

1

2

1

2 ( 1) x dx

Take out common factor


12

1

2 ( 1) x dx

13

1

23

xx

3 3(1) ( 1)2 1 ( 1)

3 3

1 12 1 1)

3 3

22 2

3

8

3

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Higher

2If f '( ) 3 4 1 and f (2) 11, find f ( ).x x x 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 11C

f

8 8 2 11 C

3 C

3 2( ) 2 3f x x x x Hence:


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Higher


5

12x dx

52

1

2

2

x

2 25 1 24

2 3

14x dx

24

2

4

4

x

4 4 2 ( 2) 6

Example :

52

1x

24

2x

Calculus Revision

Back NextQuit

Integrate 2 4 3x x dx 3 24

33 2

x xx c

3 212 3

3x x x c

Integrate term by term

simplify

Calculus Revision

Back NextQuit

Integrate 33 4x x dx4 23 4

4 2

x xc

4 23

42x x c


Calculus Revision

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Evaluate4

1

x dx1

2

4

1

x dx 43

2

1

2

3x

4

1

32

3x

3 32 24 1

3 3

16 2

3 3

14

3

2

34

Straight line form

Calculus Revision

Back NextQuit

Evaluate

2

21

dx

x2

2

1

x dx 21

1x 1 12 1

11

2

1

2

Straight line form

Calculus Revision

Back NextQuit

Integrate3

2x dxx

1

32 2x x dx 3 222

3

2

2

xx c

3222

3x x c

Straight line form

Calculus Revision

Back NextQuit

Integrate3 1x dx

x

13 2x x dx

14 2

1

24

x xc

14 21

42x x c

Straight line form

Calculus Revision

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Integrate

3 5x xdx

x

1 1

2 2

3 5x xdx

x x

5 1

2 25x x dx 7 3

2 22

7

10

3cx x

Straight line form

Calculus Revision

Back NextQuit

Integrate

324

2

x xdx

x

3

2

1 1

2 2

4

2 2

x xdx

x x

1

2 1

22x x dx

3

2 24 1

3 4x x c

Split into separate fractions

Calculus Revision

Back NextQuit

Integrate2 5x

dxx x

3

2

2 5xdx

x

3 3

2 2

2 5xdx

x x

1 3

2 25x x dx

3 1

2 2

3 1

2 2

5x xc

3 1

2 22

310x x c

Straight line form

Calculus Revision

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Find p, given

1

42p

x dx 1

2

1

42p

x dx 3

2

1

2

342

p

x

3 3

2 22 2

3 3(1) 42p

2 233 3

42p 32 2 126p

3 64p 3 2 1264 2p 1

12 32 16p

Calculus Revision

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Integrate (3 1)( 5)x x dx 23 14 5x x dx

3 23 145

3 2

x xx c

3 27 5x x x c

Multiply out brackets


simplify

Calculus Revision

Back NextQuit

Integrate2(5 3 )x dx

3(5 3 )

3 3

xc

31

9(5 3 )x c

Standard Integral

(from Chain Rule)

Calculus Revision

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Integrate 2 2

2

2 2, 0

x xdx x

x

4

2

4xdx

x

4

2 2

4xdx

x x 2 24x x dx

3 14

3 1

x xc

31

3

4x c

x

Split into

separate fractions

Multiply out brackets

Calculus Revision

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Evaluate22

2

1

1x dxx

1

222

1x x dx

Cannot use standard integral

So multiply out

4 22

12x x x dx

25 2 1

1

1

5x x x 5 2 5 21 1 1 1

5 2 5 12 2 1 1

32 1 1

5 2 54 64 40 20 2

10 10 10 10 82

10 1

58

Calculus Revision

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The graph of

32

1 1

4

dyx

dx x

( )y g x passes through the point (1, 2).

express y in terms of x. If

3 2 1

4

dyx x

dx

4 1 1

4 1 4

x xy x c

simplify

4 1 1

4 4

xy x c

x Use the point

41 1 1

2 14 1 4

c

3c Evaluate c41 1

4 4

13y x x

x

Calculus Revision

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A curve for which 26 2

dyx x

dx passes through the point (–1, 2).

Express y in terms of x.

3 26 2

3 2

x xy c 3 22y x x c

Use the point3 22 2( 1) ( 1) c 5c

3 22 5y x x

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Higher

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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Higher

Examples:

5

1(3 )x dx

52

1

32

xx

25 115 3

2 2 24

2 2

0(3 1)x dx

23

0x x 3 32 2 0 0 6


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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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Higher

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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Higher

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 44 3

2

2 2

2(2 4) 4

3

xx dx x

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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Higher

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

w

s

xA dx

Area under a Curve

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Higher

6 2

2

(9 )4w

xA dx

63

2

(9 )12

xx

218

3

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

2 2082(16 18 )

3 3A

The cargo volume is:3208

60 60. 20.208 413

60A m


3 36 2(9 6 ) (9 2 )

12 12

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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

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Higher Outcome 2

Are you on Target !

• Update you log book

• Make sure you complete and correct

ALL of the Integration questions in

the past paper booklet.

higher higher unit 2 what is integration the process of integration ( type 1 ) area between to...

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