ADDITIONAL MATHEMATICS FORM 5 - Penditamuda's Blog · PDF file3.12.2008 ·...

INTEGRATION

ADDITIONAL MATHEMATICSFORM 5

MODULE 4

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1

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



2

Indefinite Integral

Integration ofAlgebraic

Functions

c


http://mathsmozac.blo

) .= +òa a dx ax c

1

) .1

+

= ++òn

n xb x dx c

n

1

) .1

+

= = ++ò ò

nn n a x

a x d x a x d x cn

[ ]) ( ) ( ) ( ) ( )± = ±ò ò òd f x g x dx f x dx g x dxe) The Equation of a Curve fromFunctions of Gradients

'( )

( ) ,

=

= +

òy f x d x

y f x c

gspot.com

INTEGRATION

1. Integration is the reverse process of differentiation.

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

f xdx

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

If ( ), then ( )= =òdy

f x f x dx ydx

4.1. Integration of Algebraic Functions

Indefinite Integral


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

1

) .1

+

= ++òn

n xb x dx c

nc is constant, n is an integer and n ≠ -

1

) .1

+

= = ++ò òn

n n axc ax dx a x dx c

na and c are constants n is an

3

[ ]) ( ) ( ) ( ) ( )± = ±ò ò òd f x g x dx f x dx g x dx


Find the in

Solution :

2

4

3)

x xa

x

)

)

a

c

a)

x


Alwaysremember to

include ‘+c’ inyour answersof indefinite

integrals.

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

22

5 1

xdx x c b x dx c

xc

xx dx c

2 2

6 2 3

6

) ( 3 ) 3

2 3= =

6 2 3

1=

3

d x x dx xdx x dx

x x xc c

x c

23=

2

xx c

definite integral for each of the following.

2

4 4

3 2

2 1

2

3

3

= 32 1

1 3=

2

x xdx dx

x x

x x dx

x xc

cxx

2

2

4 4

3 4b) 3

xdx x dx

x x

http://mathsm
4
2 2

4 2

2 2

4 4) 3 = 3

= 3 4

b x dx x dxx x

x x dx

3 1

3

3= 4

3 1

4=

x xc

x cx

ozac.blogspot.com

5

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

3. Find2

12 .x dx

x

[3m] 4. Find 3

4

32 2 .x x dx

x

[3m]

5. Integrate3

6 5x

x

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

5 2

4

4

2

x xdx

x

[3m]

7. Find 2

6

36x dx

x

. [3m] 8. Integrate

2 3 2

1

x x

x

with respect to x.

[3m]



The Equation of a Curve from Functions of Gradients

Find the eqthe point (2

Solution

The

If th


e 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.

uation of the curve tha, −3).

gradient function is 3

3

(3

3

2

dy

dx

y

xy

ThecurveThus, x =

2

3(2)3

2

3 6 4

5

Hence, th

3

2

c

xy

http://m
6
t has the gradient function 3x − 2 and passes through

x − 2.

2

2

2)

2

x

x dx

x c

passes throughthepoint (2,−3).2, y=−3.

2

2

e equation of curve is

2 5

x c

c

x

athsmozac.blogspot.com

7

1. Given that 6 2dy

xdx

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

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

3. The gradient function of a curve is given by3

48dykx

dx 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.



8

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]



9

SPM 2005- Paper 2 :Question 2

A curve has a gradient function 2 4px 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]



1

(

(

2


10

. Find the indefinite integral for each of the following.

a) 3

2 3

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

x x

c)

22

5

5 2

32 1(c) x + (d)

3 6x

xdx dx

x

. If 34 4 ,dy

x xdx

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


11

3. If3

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 the

origin.

5. Given that2

24 ,

d yx

dx and that 0,

dy

dx y = 2 when x = 0. Find

dy

dxand y in terms

of 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

EXERCISE A

3 2

43

3

4 2

3

2

2

3

3

2

1) 2 10

2) 32

4 13) 4

3

14) 2

2 2

6 55)

2

26)

4

17) 2

8) 22

x x x c

xx x c

x x cx

x xx c

x

x x

xc

x

x cx

xx c

EXERCISE B

2

2

2

2

1) 3 2 1

2) 2 3

3 243) 2

2

y x x

y x x

xy

x

12http://mathsmozac.

ASSESSMENT

4 2

2

6

4

3

4 2

3 2

3

31) ( ) 2

2

2 3( ) 3

1( )

9 24

9( ) 6

3

2) 2 8

73)

8

2 34)

3 2

25) 2

3

a x x x c

b x cx x

xc c

x

xd x c

x

y x x

p

y x x x

y x

blogspot.com

13

INTEGR

ADDITIONAL MFOR

MOD


http://mathsmoza

ATHEMATICSM 5

ATION

ULE 5

c.blogspot.com

14

CONTENT

CONCEPT MAP 2

INTEGRATION BY SUBSTITUTION 3

DEFINITE INTEGRALS 5

EXERCISE A 6

EXERCISE B 7

ASSESSMENT 8

SPM QUESTIOS 9

ANSWERS 10



15

CONCEPT MAP

INTEGRATION BYSUBSTITUTION

n

n uax b dx du

a

where u = ax + 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

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



INTEGRATION BY SUBSTITUTION

n

n uax b dx du

a

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

1

,1

nn ax b

ax 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) 3

2 1x dx

(b) 74(3 5)x dx

16

(c)3

2

(5 3)dx

x

SOLUTION

(a) 3

2 1x 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

dux dx u

udu

uc

uc

xc

Substitute 2x+1 andsubstitute dx with

dx =2

du

Substituteu = 2x +1

OR


43

4

(2 1)(2 1)

2(4)

2 1=

8

xx dx c

xc

(b)7

77

8

8

8

4(3 5)

Let 3 5

33

44(3 5)

3

4=

3(8)

=6

(3 5)=

6

x dx

u x

du dudx

dx

ux dx du

uc

uc

uc

OR

87

8

4(3 5)4(3 5)

3(8)

(3 5)=

6

xx dx c

xc

(c)

3

3

33

3

2

2

22(5 3)

(5 3)

Let 5 3

55

22(5 3)

5

2=

5( 2)

=5

1=

5

dx x dxx

u x

du dudx

dx

ux dx du

uc

uc

u

2

1=

5(5 3)c

x

DEFINITE INTEGRALS

(a)

( )

( )

b

a

b

a

b

a

b

c


If ( ) ( ) thend

g x f xdx

17

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )

b

a

b

a

c c

b a

f x dx g x g b g a

f x dx f x dx

f x dx f x dx f x dx


18

Evaluate each of the following

(a) 21 4

( 3)( 3)x xdx

x

(b) 10 2

1

(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

xdx

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 8

1= 2

8

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 dxx

x dx

x

x

1 1=

6 2

1=

3

SOLUTION



19

INTEGRATE THE FOLLOWING USING SUBSTITUTION METHOD.

(1) 3( 1)x dx (2) 5

4 3 5x dx

(3)

4

1

5 3dx

x

(4)

31

52

x dx

(5)4

15 4

2y dy

(6)

53 2

52 3

u du

EXERCISE A



20

1. Evaluate 8 33 ( 4)x dx

Answer : 1023.752. Evaluate dxxxx )5(

8

1 24

3

Answer:96

833

3. Integrate4

25

3x

with respect to x

Answer:5

3 25

10 3x c

4. Evaluate1

41

12 3x dx

x

Answer :1

33

5. Evaluate 3

21

2 1 2 1

4

x xdx

x

Answer:5

16

6.Given that5

2( ) 10f x dx , find the value

of 2

51 2 ( )f x dx

Answer :17

EXERCISE B



21

1. Given that2

1( ) 3f x dx and

2

3( ) 7f x dx . Find

(a) 2

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

(b) 3

15 ( ) 1f x dx

Answer : (a) k =22

3(b) 48

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

(b)

5

4

3 1 5dx

x

3. Show that

2 2

2

2 6

3 2 3 2

d x x x

dx x x

.

Hence, find the value of

1

20

3

3 2

x xdx

x

.

Answer :1

10

4. Given that4

0( ) 3f x dx and

4

0( ) 5g 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



22

SPM 2003 – PAPER 1, QUESTION 17

1.Given that

4

51

1

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 , where

k > -1 , find the value of k. [4 marks]

Answer: k = 5


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

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

Answer: k =1

4

SPM QUESTIONS



23

EXERCISE A

1. 3 ( x + 1)4 + c

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

3.

3

20

5 3c

x

4.2

3 15

2 2x c

5.4

110 4

2y c

6.

62

5 53

u c

EXERCISE B

1. 1023.75

2.

833

96

3.5

3 25

10 3x c

4.1

33

5.5

16

6. 17

ASSESSMENT

1. (a) k =22

3(b) 48

2. (a) 90(2 – 3v)-5 +c

(b) 4100(1 5 )

3x c

3.1

10

4. (a) – 15(b) 4

SPM QUESTIONS

1. k =5

3 n = - 3

2. k = 5

3. k =1

4

ANSWERS



2

INTEGR

ADDITMATHE

MOD


http://mathsmoz

IONALMATICS

4

ATION

ULE 6

ac.blogspot.com

25

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



26

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

b

ay dx

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

b

ax dy

c) The area enclosed by acurve and a straight line

( ) ( )b

af x g x dx

a) The volume generated whena curve is rotated through360º about the x-axis is

2b

x aV y dx

b) The volume generated whena curve is rotated through360º about the y-axis is

2b

ya

V x dy



27

3. INTEGRATION

3.1 Integration as Summation of Area

The area under a curve which enclosed by

x = a and x = b isb

aydx

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

The area under a curve which is enclosed

by y = a and y = b isb

axdy

Note :The area is preceded by a negative sign ifthe 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 af x dx g x

= ( ) ( )b

af x g x dx

xa b

y = f(x)

0

x

y = f(x)

b

a

0

y

x

y = g (x)

y = f (x)

a b

y



28

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

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

y = x2 – 2x

x

y

0

y = -x2 + 3x+ 4

x-1 40

y



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

y = 2

y

x

y = x2 + 4x + 4

y


0
x = y2
http://mathsm
29
-1-2x

20

ozac.blogspot.com

5. Find the area of the shaded region in thediagram

6.

x = y3 - y

1

y

x

y y = ( x – 1)2


0

-1

Given that the area of the shaded region in

x = kx

0

http://m
30
the diagram above is2

3value of k.

athsmozac.blogspot.com

8units2. F
ind the

3.2 Integration as Summation of Volumes

y

x

y=f(x)

0 a b

The volume generatedwhen a curve is rotatedthrough 360º about thex-axis is

2b

x aV y dx


3

The volume generatedwhen a curve is rotatedthrough 360º about they-axis is

2b

ya

V x dy

y

y=f(x)

a

b

0

http://mathsmoz

x

1ac.blogspot.com

Answer :

Answer :

Given

substitute

Then,

y

x

y

y


2

y

x

x=2

y=x(x+1)

0

Find the volume generated when theshaded region is rotated through 360ºabout the x-axis.

2

0

2 22

0

24 3 2

0

25 4 3

0

5 4 3

3

1

( 2 )

2

5 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

Volume generated

y

0

26

0 int

6 0

6

x

62

x

26y x The figure shows the shaded region that is

enclosed by the curve 26y x , the

x-axis and the y-axis.

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

32

0

6

0

62

0

2

3

6

62

66(6) 0

2

18 .

x dy

y dx

yy

units

Volume generated

2o 6y x


33

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 360º about the y-axis. [4 marks]

y = x (2 – x)

0

y

x



34

2.

The figure shows the curve 2( 2)y x . Calculate the volume generated when

the shaded region is revolved through 360º about the x-axis.

y² = 4 (x + 1)

0x

y

Q (3, 4)

P (0, 2)

R (0, 4)

x=3



3

3.

The above figure shows part of

x = k. If the volume generated

360º about the x-axis is1

122

0x

y

x = k

R (0, 4)


http://mathsmoz
5
the curve 3y x and the straight line

when the shaded region is revolved through

3units , find the value of k.

3y x

ac.blogspot.com

36

SPM 2003- Paper 2 :Question 9 (b)

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

3 2y x

Diagram 3

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

y

x

3 2y x

2 1x y

−1 0



37


Diagram 5 shows part of the curve

2

3

2 1y

x

which 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 360º about the x-axis.

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

2

3

2 1y

x

(1,3)A

y

x0




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

12

xy at A(2, 3). The

straight line AR is parallel to the y-axis.

DFind(a) the value of k,(b) the area of the sh(c) the volume gene

y-axis and the str

y

x

A(2, 3)

Q(k, 0)0

2

12

xy


ia

adratai

R

h
38
gram 4

[3 marks]ed region, [4 marks]ed, in terms of , when the region bounded by the curve, theght line y = 3 is revolved through 360º about y-axis. [3 marks]

ttp://mathsmozac.blogspot.com

EXER

1.

2.

3.

4.

5.

6.


39

CISE A

2

2

2

2

2

11 units

3

520 units

6

22 units

3

224 units

3

1units

2

4k

http://mathsmozac.

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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ADDITIONAL MATHEMATICS FORM 5 - Penditamuda's Blog · PDF file3.12.2008 ·...

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Transcript of ADDITIONAL MATHEMATICS FORM 5 - Penditamuda's Blog · PDF file3.12.2008 ·...