additional mathematics form 5 - penditamuda's blog · pdf file3.12.2008 ·...
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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
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2
Indefinite Integral
Integration ofAlgebraic
Functions
c
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) .= +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
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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
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) .= +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
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Find the in
Solution :
2
4
3)
x xa
x
)
)
a
c
a)
x
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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
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42 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
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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]
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The Equation of a Curve from Functions of Gradients
Find the eqthe point (2
Solution
The
If th
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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
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6t 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
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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.
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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]
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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]
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1
(
(
2
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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.
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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.
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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
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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
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INTEGR
ADDITIONAL MFOR
MOD
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ATHEMATICSM 5
ATION
ULE 5
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CONTENT
CONCEPT MAP 2
INTEGRATION BY SUBSTITUTION 3
DEFINITE INTEGRALS 5
EXERCISE A 6
EXERCISE B 7
ASSESSMENT 8
SPM QUESTIOS 9
ANSWERS 10
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
SPM 2004 – PAPER 1, QUESTION 22
2. Given that 1 2 3 6k x dx , where
k > -1 , find the value of k. [4 marks]
Answer: k = 5
SPM 2005 – PAPER 1, QUESTION 21
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
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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 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
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2
INTEGR
ADDITMATHE
MOD
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IONALMATICS
4
ATION
ULE 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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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
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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
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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
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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
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0
x = y2http://mathsm
29-1-2x
20
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5. Find the area of the shaded region in thediagram
6.
x = y3 - y
1
y
x
y y = ( x – 1)2
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0
-1
Given that the area of the shaded region in
x = kx
0
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30the diagram above is2
3value of k.
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8units2. F
ind the3.2 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
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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
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Answer :
Answer :
Given
substitute
Then,
y
x
y
y
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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
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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
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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
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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)
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5the curve 3y x and the straight line
when the shaded region is revolved through
3units , find the value of k.
3y x
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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
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37
SPM 2004- Paper 2 :Question 10
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
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SPM 2005- Paper 2 :Question 10
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
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ia
adratai
R
h
38gram 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]
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EXER
1.
2.
3.
4.
5.
6.
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39
CISE A
2
2
2
2
2
11 units
3
520 units
6
22 units
3
224 units
3
1units
2
4k
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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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