Integration

Higher Mathematics

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Calculus Revision

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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

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Find 3cos x dx3sin x c

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

Integrate term by term

simplify

Calculus Revision

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Find 2sin d 2cos c

Calculus Revision

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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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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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Evaluate

2

21

dx

x2

2

1

x dx 21

1x 1 12 1

11

2

1

2

Straight line form

Calculus Revision

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Find

1

6

0

(2 3)x dx17

0

(2 3)

7 2

x

7 7(2 3) (0 3)

14 14

7 75 3

14 14

5580.36 156.21 (4sf)5424

Use standard Integral(from chain rule)

Calculus Revision

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Find1

3sin cos2

x x dx1

3cos sin2

x x c Integrate term by term

Calculus Revision

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Integrate3

2x dxx

1

32 2x x dx 3 222

3

2

2

xx c

3222

3x x c

Straight line form

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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

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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

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Find 2

3

1

2 1x dx

24

1

2 1

4 2

x

4 44 1 2 1

4 2 4 2

4 45 3

8 8

68

Use standard Integral(from chain rule)

Calculus Revision

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Find sin 2 cos 34

x x dx

1 1

2 3cos 2 sin 3

4x x c

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Find2

sin7

t dt2

7cos t c

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Integrate

1

2

1

0 3 1

dx

x

1

2

1

0

3 1x dx

Straight line form

1

2

1

0

13

2

3 1x

1

0

23 1

3x

2 23 1 0 1

3 3

2 24 1

3 3

4 2

3 3

2

3

Calculus Revision

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Given the acceleration a is: 1

22(4 ) , 0 4a t t

If it starts at rest, find an expression for the velocity v where dv

adt

1

22(4 )dv

tdt

3

2

31

2

2(4 )tv c

3

24

(4 )3

v t c

344

3v t c Starts at rest, so

v = 0, when t = 0 340 4

3c

340 4

3c

320

3c

32

3c

3

24 32

3 3(4 )v t

Calculus Revision

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A curve for which 3sin(2 )dy

xdx

5

12, 3passes through the point

Find y in terms of x. 3

2cos(2 )y x c

3 5

2 123 cos(2 ) c Use the point 5

12, 3

3 5

2 63 cos( ) c 3 3

2 23 c

3 33

4c

4 3 3 3

4 4c

3

4c

3 3

2 4cos(2 )y x

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 intoseparate fractions

Multiply out brackets

Calculus Revision

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If

( ) sin(3 )f x x passes through the point 9, 1

( )y f x express y in terms of x.

1

3( ) cos(3 )f x x c Use the point 9

, 1

1

31 cos 3

9c

1

31 cos

3c

1 1

3 21 c

7

6c 1 7

3 6cos(3 )y x

Calculus Revision

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Integrate2

1

(7 3 )dx

x2(7 3 )x dx

1(7 3 )

1 3

xc

11

3(7 3 )x c

Straight line form

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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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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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

Calculus Revision

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Evaluate

222

1

1x dxx

1

222

1x x dx

Cannot use standard integralSo 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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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

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Evaluate0

2

3

(2 3)x dx

Use standard Integral(from chain rule)

03

3

(2 3)

3 2

x

3 3(2(0) 3) (2( 3) 3)

6 6

27 27

6 6

27 27

6 6

54

6 9

Calculus Revision

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The curve ( )y f x ,112

passes through the point

( ) cos 2f x x Find f(x)

1

2( ) sin 2f x x c

1

2 121 sin 2 c

use the given point ,112

1

2 61 sin c

1 1

2 21 c 3

4c 1 3

2 4( ) sin 2f x x

1

6 2sin

Calculus Revision

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2(6 cos )x x x dx Integrate

3 26sin

3 2

x xx c Integrate term by term

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Integrate 33 4x x dx4 23 4

4 2

x xc

4 23

42x x c

Integrate term by term

Calculus Revision

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Evaluate1

01 3x dx

1

21

01 3x dx

13

2

0

3

2

1 3

3

x

13

2

0

21 3

9x

13

0

21 3

9x

3 32 21 3(1) 1 3(0)

9 9

3 32 24 1

9 9

16 2

9 9

14

9 5

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