Integration Modul 4 PDF December 3 2008-1-05 Am 640k

8/12/2019 Integration Modul 4 PDF December 3 2008-1-05 Am 640k

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INTEGRATION

ADDITIONAL MATHEMATICS

FORM 5

MODULE 4

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CHAPTER 3 : INTEGRATION

Content page

Concept Map 2

4.1 Integration of Algebraic Functions 3 4

Exercise A 5

4.2 The Equation of a Curve fromFunctions of Gradients. 6

Exercise B 7

SPM Question 8 9

Assessment 10 11

Answer 12




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2

Indefinite Integral

) .= +a a dx ax c

Integration ofAlgebraicFunctions

1) .

1+

= ++nn xb x dx c

n

1

) .1

+

= = ++

nn n a xc a x d x a x d x c

n

[ ]) ( ) ( ) ( ) ( = d f x g x dx f x dx g xe) The Equation of a Curve fromFunctions of Gradients

'( )

( ) ,

=

= + y f x d x

y f x c




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3

INTEGRATION

1. Integration is the reverse process of differentiation .

2. If y is a function of x and '( )=dy

f xdx

then '( ) , constant.= + = f x dx y c c

If ( ), then ( )= =dy

f x f x dx ydx

4.1. Integration of Algebraic Functions

I ndefini te I ntegral

) .= +a a dx ax c a and c are constants

1) .

1

+= +

+nn xb x dx c

n c is constant, n is an integer and n -

1

) .1

+

= = ++ n

n n axc ax dx a x dx cn

a and c are constants n is an

[ ]) ( ) ( ) ( ) ( ) = d f x g x dx f x dx g x dx




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4

Find the indefinite integral for each of the following.

Solution :

2 2

4 4 4

3 2

2 1

2

3 3)

3

= 32 1

1 3=

2

x x x xa dx dx

x x x

x dx

x xc

c x x

Find the indefinite integral for each of the following.3

5 2

) 5 )

) 2 ) ( 3 )

a dx b x dx

c x dx d x x dx

Solution :

3 13

4

5 15

) 5 5 )3 1

=4

2) 25 1

xa dx x c b x dx c

xc

xc x dx c

2 2

6 2 3

6

) ( 3 ) 3

2 3= =

6 2 31

=3

d x x dx xdx x dx

x x xc c

x c

23=

2 x

x c

2 24 4

3 4a) b) 3

x xdx x dx

x x

Always remember to

include +c in

your answers of ind ef in ite integrals.

2 24 2

2 2

4 4) 3 = 3

= 3 4

b x dx x dx x x

x x dx

3 1

3

3= 4

3 1

4=

x xc

x c x




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5

1. Find 23 4 10 . x dx [3m] 2. Find 2 1 2 3 . x x dx [3m]

3. Find2

12 .dx

x

[3m] 4. Find 3 43

2 2 . x dx x

[3m]

5. Integrate 36 5 x

xwith respect to x. [3m] 6. Find

5 24

4

2

x xdx

x [3m]

7. Find 2 63

6 x dx x

. [3m] 8. Integrate2 3 2

1 x x

x

with respect to x.

[3m]




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6

The Equation of a Curve from Functions of Gradients

Find the equation of the curve that has the gradient function 3 x 2 and passes throughthe point (2, 3).

Solution

The gradient function is 3 x 2.

2

3 2

(3 2)

3 22

dy x

dx

x dx

x x c

Thecurve passes throughthepoint (2, 3).Thus, x =2, y = 3.

2

2

3(2)3 2

23 6 4

5

Hence, the equation of curve is

32 5

2

x c

c

c

x y x

If the gradient function of the curve is '( ),dy

f xdx

then the equation of the curve is

'( )

( ) ,

y f x dx

y f x c c is constant.




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1. Given that 6 2dydx

, express y in terms of x if y = 9 when x = 2.

2. Given the gradient function of a curve is 4 x 1. Find the equation of the curve if it passes through the point ( 1, 6).

3. The gradient function of a curve is given by 348dy

kxdx x

, where k is a constant.

Given that the tangent to the curve at the point (-2, 14) is parallel to the x-axis, findthe equation of the curve.




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SPM 2003- Paper 2 :Question 3 (a)

Given that 2 2dy

xdx

and y = 6 when x = 1, find y in terms of x. [3 marks ]

SPM 2004- Paper 2 :Question 5(a)

The gradient function of a curve which passes through A(1, 12) is 23 6 . x x Find theequation of the curve. [ 3 marks ]




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SPM 2005- Paper 2 :Question 2

A curve has a gradient function 2 4 px x , where p is a constant. The tangent to the curve

at the point (1, 3) is parallel to the straight line y + x 5 =0.Find(a) the value of p, [3 marks ](b) the equation of the curve. [3 marks ]




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1. Find the indefinite integral for each of the following.

(a)

3

2 3

2 64 3 2 (b) 3 x x dx dx

x x

( c) 22

55 2

32 1(c) x + (d)3 6x

xdx dx x

2. If 34 4 ,dy

x xdx

and y = 0 when x = 2, find y in terms of x.




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3. If 3

2 ,2

dp vv

dv and p = 0 when v = 0, find the value of p when v = 1.

4. Find the equation of the curve with gradient 22 3 1, x x which passes through theorigin.

5. Given that2

2 4 ,d ydx

and that 0,dydx

y = 2 when x = 0. Find dydx

and y in terms

of x.




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EXERCISE A

3 2

43

3

4 2

3

2

2

33

2

1) 2 10

2) 32

4 13) 4

31

4) 22 2

6 55)

2

26)4

17) 2

8) 22

x x x c x

x x c

x x c x

x x x c

x

x x

x c x

x c x

x x c

EXERCISE B

2

2

2

2

1) 3 2 1

2) 2 3

3 243) 2

2

y x x

y x x

x y

x

SPM QUESTIONS

2

3 2

3 2

1) 2 7

2) 3 10

3) 3,

2 4

y x x

y x x

p

y x x

ASSESSM EN T

4 2

2

6

4

3

4 2

3 2

3

31) ( ) 2

22 3

( ) 3

1( )

9 249

( ) 63

2) 2 8

73)

8

2 34)

3 2

25) 23

a x x x c

b x c x x

xc c

x x

d x c x

y x x

p

y x x x

y x




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INTEGRATION

ADDITIONAL MATHEMATICS

FORM 5

MODULE 5




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O N T N T

CONCEPT MAP 2

INTEGRATION BY SUBSTITUTION 3

DEFINITE INTEGRALS 5

EXERCISE A 6

EXERCISE B 7

ASSESSMENT 8

SPM QUESTIOS 9ANSWERS 10




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CONCEPT MAP

INTEGRATION BYSUBSTITUTION

n

n uax b dx dua

where u = a x + b,a and b are constants , n is aninteger and n -1

OR

1

,1

nn ax b

ax b dx ca n

where a, b, and c are constants , n isinteger andn -1

DEFINITE INTEGRALS

If ( ) ( ) thend

g x f xdx

(a) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

bb

a a

b b

a a

b c c

a b a

x dx g x g b g a

b f x dx f x dx

c f x dx f x dx f x dx




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INTEGRATION BY SUBSTITUTION

nn uax b dx du

a

where u = ax + b,a and b areconstants, n is aninteger and n -1

1

,1

n

n ax bax b dx ca n

where a, b, and c areconstants, n is integer andn -1

OR

Find the indefinite integral for each of the following.

(a) 32 1 x dx (b) 74(3 5) dx

(c) 32

(5 3)dx

x

SOLUTION

(a) 3

2 1 dx

Let u = 2x +1

22

du dudx

dx

3 3

3

3 1

4

(2 1)2

=2

=2(3 1)

= +8

(2 1)= +

8

du x dx u

udu

u c

uc

xc

Substitute 2 x+1 andsubstitute dx with

dx =2

du

Substituteu = 2x +1

OR

43

4

(2 1)(2 1)

2(4)

2 1=

8

xdx c

xc




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(b)7

77

8

8

8

4(3 5)

Let 3 5

3

34

4(3 5)3

4=

3(8)

=6(3 5)

=6

x dx

u x

du dudx

dxu

x dx du

uc

uc

uc

OR

87

8

4(3 5)4(3 5)

3(8)

(3 5)=

6

xdx c

xc

(c)3

3

33

3

2

2

22(5 3)

(5 3)

Let 5 3

5 52

2(5 3)5

2=

5( 2)

=5

1=

5

dx x dx x

u x

du dudxdxu

x dx du

uc

uc

u

2

1=

5(5 3)c

x

DEFINITE INTEGRALS

If ( ) ( ) thend

g x f xdx

(a) ( ) ( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( )

bb

a a

b b

a a

b c c

a b a

f x dx g x g b g a

b f x dx f x dx

c f x dx f x dx f x dx




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Evaluate each of the following

(a) 21 4( 3)( 3) x x dx

(b) 10 21

(2 1)dx

x

(a)2

2 21 14 4

22

1 4 4

2 2 41

21 3

1

( 3)( 3) 9

9=

= ( 9 )

= 91 3

x x xc dx

x x

x dx x x

x x dx

x x

2

31

3 3

1 3=

1 3 1 3= 2 2 1 1

1 3= ( 1 3)

2 81

= 28

1= 2

8

x x

(b)

1 1 20 02

1 20

11

0

1

0

1(2 1)

(2 1)

= (2 1)

(2 1)=1(2)

1=

2(2 1)

1 1=

2[2(1) 1] 2[2(0) 1]

dx x dx x

x dx

x

x

1 1

=6 2

1=3

SOLUTION




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INTEGRATE THE FOLLOWING USING SUBSTITUTION METHOD.

(1) 3

( 1) dx (2) 5

4 3 5 dx

(3) 4

1

5 3dx

x (4)

315

2 x dx

(5)415 4

2dy

(6)53 25

2 3u du

EXERCISE A




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1. Given that2

1( ) 3 f x dx and

2

3 ( ) 7 f x dx . Find

(a) 2

1the value of k if ( ) 8kx f x dx

(b) 3

15 ( ) 1 f x dx

Answer : (a) k = 22

3(b) 48

2.(a) 65(2 3 )v dv

(b) 54

3 1 5 dx x

3. Show that

2 2

22 63 2 3 2d x x xdx x x

.

Hence, find the value of

1

20

3

3 2

x xdx

x .

Answer : 110

4. Given that4

0 ( ) 3 f x dx and4

0( ) 5 g x dx . Find

(a)4 0

0 4( ) ( ) f x dx g x dx

(b) 4

03 ( ) ( ) f x g x dx

Answer: (a) 15(b) 4

ASSESSMENT




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SPM 2003 PAPER 1, QUESTION 17

1.Given that

45

11

ndx k x c

x ,

find the value of k and n [3 marks ]

Answer: k = 5

3 n = - 3


2. Given that 1 2 3 6k x dx , wherek > -1 , find the value of k. [4 marks ]

Answer: k = 5


3. Given that 62 ( ) 7 f x dx and 6

2 (2 ( ) ) 10 f x kx dx , find the value of k.

Answer: k = 14

SPM QUESTIONS




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EXERCISE A

1. 3 ( x + 1)4 + c

2. 60 (3 x +5) - 4 + c

3. 3

20

5 3c

x

4.2

3 152 2

x c

5.4

110 4

2c

6.

62

5 53

u c

EXERCISE B

1. 1023.75

2.

833

96

3.5

3 25

10 3c

4. 13

3

5. 51

66. 17

ASSESSMENT

1. (a) k = 22

3(b) 48

2. (a) 90(2 3v) -5 +c

(b) 4100

(1 5 )3

c

3. 110

4. (a) 15(b) 4

SPM QUESTIONS

1. k = 5

3 n = - 3

2. k = 5

3. k = 14

ANSWERS




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INTEGRATION

ADDITIONALMATHEMATICS

MODULE 6




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CHAPTER 3 : INTEGRATION

Content page

Concept Map 2

9.1 Integration as Summationof Areas 3

Exercise A 4 6

9.2 Integration as Summationof Volumes 7 8

Exercise B 9 11

SPM Question 12 14

Answer 15




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a) The area under a curvewhich enclosed by x -axis, x = a and x = b is

b

a y dx

b) The area under a curvewhich enclosed by y -axis, y = a and y = b is

b

a x dy

c) The area enclosed by acurve and a straight line

( ) ( )b

a f x g x dx

a) The volume generated whena curve is rotated through360 about the x -axis is

2b

x aV y dx

b) The volume generated whena curve is rotated through360 about the y -axis is

2b

y aV x dy




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3. INTEGRATION

3.1 Integration as Summation of Area

The area under a curve which enclosed by

x = a and x = b is ba

dx

Note : The area is preceded by a negative sign if the region lies below the x axis.

The area under a curve which is enclosed by y = a and y = b is

b

a xdy

Note : The area is preceded by a negative sign if the region is to the left of the y axis.

The area enclosed by a curve and a straight line

The area of the shaded region = ( ) ( )b b

a a f x dx g x

=

( ) ( )b

a f x g x dx

xa b

y = f(x)

0x

y = f(x)

b

a

0

y

x

y = g (x)

y = f (x)

a b

y




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1. Find the area of the shaded region inthe diagram.

2. Find the area of the shaded region in thediagram.

y = x 2x

x

y

0

y = -x2 + 3x+ 4

x-1 40

y




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3. Find the area of the shaded region 4. Find the area of the shaded region in thediagram.

y = 2

0 x

x = y -1-2

y = x + 4x + 4

x

y

20




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5. Find the area of the shaded region in thediagram

6.

Given that the area of the shaded region in

the diagram above is 28

3units 2. Find the

value of k.

x = y - y

0

1

-1

xx = k

y

x0

y = ( x 1)




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3.2 Integration as Summation of Volumes

y

x

y=f(x)

0 a b

The volume generated when a curve is rotated through 360 about the x-axis is

2b

x aV y dx

The volume generated when a curve is rotated through 360 about they-axis is

2b

y aV x dy

y

x

y=f(x)

a

b

0




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Answer :

2 2

0

2 22

0

2 4 3 2

0

25 4 3

0

5 4 3

3

1

( 2 )

25 4 3

2 2(2) 20

5 4 3

256 1@ 17 .

15 15

y dx

x x dx

x x x dx

x x x

units

y

x

x=2

y= x( x+1)

0

Volume generated

Find the volume generated when theshaded region is rotated through 360

about the x-axis.

Answer :

62

0

6

0

62

0

2

3

6

62

66(6) 0

2

18 .

x dy

y dx

y y

units

y

x

26 x

0

Volume generated

The figure shows the shaded region that isenclosed by the curve 26 y x , the x-axis and the y-axis.

Calculate the volume generated when theshaded region is revolved through 360about y-axis.

2

2

Given 6

substitute 0 into 6

Then, 6 0

6

y x

y x

y

y




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1.

The above figure shows the shaded region that is enclosed by the curve y = x (2 x)and x-axis. Calculate the volume generated when the shaded region is revolvedthrough 360 about the y-axis. [4 marks ]

y = x (2 x)

0

y

x




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2.

The figure shows the curve 2( 2) y x . Calculate the volume generated whenthe shaded region is revolved through 360 about the x-axis.

y = 4 ( x + 1)

0 x

y

Q (3, 4)

P (0, 2)

R (0, 4)

x=3




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3.

The above figure shows part of the curve 3 y x and the straight line x = k . If the volume generated when the shaded region is revolved through

360 about the x-axis is 31

12 units ,2

find the value of k .

3 x

0 x

y

x = k

R (0, 4)




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SPM 2003- Paper 2 :Question 9 (b)

Diagram 3 shows a curve 2 1 x y which intersects the straight line 3 2 y x at point A.

3 2 y x

Diagram 3

Calculate the volume generated when the shaded region is involved 360 about the y-axis.[6 marks ]

y

x

3 2 y x

2 1 x y

1 0




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Diagram 5 shows part of the curve 2

3

2 1 y

xwhich passes through A(1, 3).

Diagram 5

a) Find the equation of the tangent to the curve at the point A.[4 marks ]

b) A region is bounded by the curve, the x-axis and the straight lines x= 2 and x= 3.i) Find the area of the region.ii) The region is revolved through 360 about the x-axis.

Find the volume generated, in terms of .[6 marks ]

23

2 1 y

x

(1,3) A

y

x0




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In Diagram 4, the straight line PQ is normal to the curve2

12

x y at A(2, 3). The

straight line AR is parallel to the y-axis.

Diagram 4Find(a) the value of k , [3 marks ](b) the area of the shaded region, [4 marks ](c) the volume generated, in terms of , when the region bounded by the curve, the

y-axis and the straight line y = 3 is revolved through 360 about y-axis. [3 marks ]

y

x

A(2, 3)

Q(k , 0)0 R

2

12

x y




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EXERCISE A

2

2

2

2

2

11. 1 units

3

52. 20 units

6

23. 2 units

3

24. 24 units

3

15. units

2

6. 4k

EXERCISE B

2

3

11. 1 unit

15

32. 6 unit

5

3. 2k

SPM QUESTIONS

3

2

3

2

2003

52units

15

2004

1)5

49)

1125

2005

) 8

1) 12

3) 4

SPM

Volume Generated

SPM

i Area units

ii Volume Generated units

SPM

a k

b Area units

c Volume Generated units


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