integration modul 4 pdf december 3 2008-1-05 am 640k
TRANSCRIPT
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INTEGRATION
ADDITIONAL MATHEMATICS
FORM 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
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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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7
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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10
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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12
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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14
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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18
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
SPM 2004 PAPER 1, QUESTION 22
2. Given that 1 2 3 6k x dx , wherek > -1 , find the value of k. [4 marks ]
Answer: k = 5
SPM 2005 PAPER 1, QUESTION 21
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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SPM 2004- Paper 2 :Question 10
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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SPM 2005- Paper 2 :Question 10
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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