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WAJA 2009
ADDITIONAL MATHEMATICS
FORM FIVE
( Student’s Copy )
Name: ___________________________
Class : ___________________________
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
3.1 Indefinite Integral
Learning Objective3.1.0 Understand and use the concept of indefinite integral.
Learning Outcome3.1.1 Determine integrals by reversing differentiation.
We can think of this as a process of reversing differentiation (anti-differentiation), it is actually called integration and the result in an integral.
is the sign of integration
Example : .
1. Fill in the blanks in the following figure.
y y = f(x)
x
y = f(x)
Gradient of tangent, m=
y
m =
x
2. Fill in the blanks in the following.
2
You have probably wondered if it is possible to find a function given its
gradient function. For example, if = 6x + 2, is it possible to find y ?
Process of
__________________
Process of
___________________
must both being written. And read as the integral of 6x + 2 with respect to x.
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
Function / Equation, y = f(x) Differentiation, Integration
y = 3x = (3x) = 3
3 dx = 3x + cy = 3x + 1
= (3x + 1) = 3
y = 3x + 2 = (3x + 2) =
y = 3x2 = (3x2) = 6x
6x dx = 3x2 + cy = 3x2 + 2 = (3x2 + 2) =
y = 3x2 + 5 = =
y = 3x4 = (3x4) = 12x3
dx = +
y = 3x 4 + 3 = (3x4 + 3) =
y = 3x 4 + 4 = =
3. In each of the following cases, the derivative of a function where c is a constant is given. Match the functions.
(4x
+ c) = 4(10x ) dx f(x) = 4x + c
(5x2
+ c) = 10x(6x -2) dx f(x) = 5x2 + c
(3x2
– 2x + c) = 6x – 24 dx f(x) = 4x 2 + 3x + c
(4x 2
+ 3x + c) = 8x + 3(8x + 3) dx f(x) = 3x2 – 2x + c
3
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
Learning Outcome3.1.2 Determine integrals of axn , where a is a constant and n is an integer, n -1. Formula for integration of axn
1. Find the indefinite integrals of the following.
a) 3 x4 dx
Solution:
=> x4 dx = 3 + c
= 3 x 5 + c 5
b) dx
Solution:
=> x -4 dx =
2. Find the indefinite integral of each of the following. (Check your answers by differentiation)
a. 2 dx =
= 2x 0 dx
= 2 + c
= 2 x 1 + c
b. 5 x 3 dx
= 5 x + c
= 5 x + c
c. 3x -2 dx
= 3 x + c
= 3 x + c
= -3 x -1 +
d. dx e. dx f. dx
4
Step 1 : Keeping the original coefficient.=> 3
Step 2 : Add 1 to the index of the term.4 + 1Step 3 : Divide the term by the new index. (4 + 1)Step 4 : Add an arbitrary constant c.
0 + 1x
3 + 1
0 + 1
Step 1 : Keeping the original coefficient.=> 2
Step 2 : Add 1 to the index of the term.-4 + 1Step 3 : Divide the term by the new index. (-4 + 1)Step 4 : Add an arbitrary constant c.
n dx = a + c , where c is a constant and n -1.
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
= x + c
= x + c
= x +
= x + c
= x + c
= x +
= dx
= dx
= x + c
= x + c
= x +
Learning Outcome 3.1.3 Determine integrals of a sum of algebraic expressions.
Formula for integration of a sum of algebraic expressions
1. Find the integral of each of the following :
a) (x2 + 3x – 2) dx
= x2 dx + 3x dx - 2 dx
= + 3( )- 2x + c
= + 3( ) - 2x + c
b) (x2 + 3x – ) dx
= (x2 + 3x –2x -2) dx
= ( ) dx + ( ) dx - ( ) dx
=
=
=
2. Arrange and rewrite the process of integration with respect to x.
Process of integration
5
[ f(x) g(x) ] dx = dx dx
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
1.
2.
3. dx + dx
4.6 x (1 + 1) + 2 x + c
(1 + 1)
3. Find the indefinite integral of each of the following. (Check your answers by differentiation)
1. (4x + 3) dx
= 4x dx + 3 dx
= 4 x 1 + 1 + 3 x 0 + 1 + c 2 1
= x + x +
2. (x3 - 4x + 2) dx
= x3 dx - 4x dx + 2 dx
= x - 4 x + 2 x + c
= x - x + x +
4. Identify the errors in the following integration solutions. Show the correct solutions.
Learning Outcome3.1.4 Find the constants of integration, c, in indefinite integrals.
Function/Equation, y = 3x2 + 2x + c
6
3x2 + 2x + c
a) (4x3 + 5x2 – 2) dx
= 4x3 dx + 5x2 dx - 2 dx
= 4x 3 + 1 + 5x 2 + 1 3+1 2+1
= 4( ) + x 3
= x4 + x
b) (4x3 + 5x2 – ) dx
= 4x3 dx + 5x2 dx - 2x -2 dx
= 4x 3 + 1 + 5x 2 + 1 – 2x -2 + c 3 + 1 2 + 1 -2
= 4x 4 + 5x 3 - 2x -2 + c 4 3 -2
= x4 + x3 + x -2 + c
cc
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
Gradient of tangent, = 6x + 2
To find the value of c, identify for the particular points on the curve (a pair of value of x and y) which is given in the question, then substitute the given information to the equation y.
1. If = 6x + 2, find the constant of integration, c, given that y = 3 and x = 1.
y =
= 6 + 2x + c
= 3x2 + 2x + c
Then, ( ) = 3( )2 + 2( ) + c
= + c
c =
2. Find the value of c.
Equation , y x y c
P(0,5) 6x + 2 y= 3x2 + 2x + c 0 5
Q(-1,6) 6x + 2 y= -1 6
R(0,-3) 6x + 2 y= 0 -3
S(-1,-2) 6x + 2 y= -1 -2
Solution :
=> =
=> y =
7
Process ofDifferentiation
Process ofIntegration
y
x
PQ(-1,6)
RS(-1,-2)
5
-3
0
c is the y-interscept of the
curve
Substitute the value y = 3 and x = 1.
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
=> At point P(0,5), x = 0 , y = 5.
c =
=> At point Q(-1,6), x = -1 , y = 6.
c =
=> At point R(0,-3), x = 0 , y = -3.
c =
=> At point P(-1,-2), x = -1 , y = -2.
c =
Learning Outcome 3.1.5 Determine equation of the curves form the functions of gradients.
1. If = 8x + 1, find the constant of integration, c, given that y = 0 and x = 1.
Hence, state the equation of the curve.
y =
y = 8 + x + c
y = 4x2 + x + c
Then, ( ) = 4( )2 + ( ) + c =
c =
The equation of the curve, y = _____________________
2. Determine the equation of the curve that has the gradient function and passes through
the point P..
a) b)The curve passing through Gradient function ,
The curve passing through Gradient function ,
8
Substitute the value y = 0 and x = 1.
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
P(0,6) x + 3 P(-2,5) 2x – 1
=> = x + 3
y = dx y = + c
Since the curve passing through P(0 , 6),substitute the value y = 6 and x = 0.
= + c
c =
Equation of the curve, y =
=> = 2x - 1
y = dx y = + c
Since the curve passing through P(-2 , 5), substitute the value y = 5 and x = -2.
= + c
c =
Equation of the curve, y =
Learning Outcome 3.1.6 Determine integrals of (ax + b)n .
Formula for integration of (ax + b)n .
1. Find the integral of each the following :
a) (3x – 2)4 dx
= (3x – 2 ) 4 + 1 + c (3)(4 + 1)
= (3x – 2 ) 5 + c 15
b) (4x + 5 dx
=
=
=
=
2. Fill in the blanks.
a) (5x – 1)2 dx b) (4x + 5)3 dx c) (3x – 7) dx
= (5x - 1) +
= (4x + 5) +
= (3x – 7) +
9
n dx = (ax + b) n+1 + c , where c is a constant and n -1. (n + 1) (a)
2 + 1
2 + 15c
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
= (5x - 1) + = (4x + 5) +
= (3x – 7) +
= (5x - 1) +
= (4x + 5) +
= (3x – 7) +
3. Match the indefinite integrals of each of the following.
1. (3x + 4)2 dx = - (2x – 1)-1 + c
2. (2x – 3)3 dx = - (5x – 1)-2 + c
3. 3(1 – 4x)5 dx = - (1 – 4x)6 + c
4. (5x – 1)-3 dx = (3x + 4)3 + c
5. 3 . dx (2x – 1)2
= (2x – 3)4 + c
3.2 Definite Integral
Learning Objective 3.2.0 Understand and use the concept of definite integral.
Learning Outcome3.2.1 Find definite integrals of algebraic expressions.
1. Evaluate each of the following.
a) x2 dx = [ ]
= [ ] – [ ]
= [ 9 ] - [ ]
b) (x2 - 3x + 1) dx
= [ - + x ]
=
=
10
3
5 3
3
15
Definite integral , y dx = [ g(x) ]
= g(b) – g(a)
a and b are known as the lower limit and upper limit
c
c
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
= =
2. Evaluate the definite integral.
(6x + 2) dx
1. [ x2 + 2x ]
2.28
3. [ 3x2 + 2x ]
4.[ 33 ] – [ 5 ]
5. [ 3(3)2 + 2(3) ] – [3(1)2 + 2(1) ]
3. Evalute dx . Draw arrow lines to join the relation according to process of
definite integral
Learning Outcome 3.2.3 Determine areas under curves using formula.
Area under the graph, y dx
11
Applications of integration
dx
= 18
=[ 2 x2 = 2(3)2 – 2(0)2
= [ 4()
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
y y = f(x)
x
Equation, y = f(x)
Function/Equation, y
Gradient of tangent, m=
y
m =
x
Further notes on areas
12
Process ofDifferentiation
Process ofIntegration
Process ofIntegration,
definite integral , y dx
a b
Area under the graph , y dx = [ g(x) ]
This is called definite integral of y respect to x between the limits a (lower limit) and b (upper limit).
a and b are known as the lower limit and upper limit
y
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
From the graph , it is clear that
f(x) dx = f(x) dx + f(x) dx
If y is negative in the range a to b, then the value obtained
form the integral f(x) dx will also be negative .
Thus, the numerical value of the area shown shaded
will be - f(x) dx or f(x) dx .
The area between a curve, y-axis, the lines
y =a and y = b will be x dy .
1. Match the following definite interal with respect to x.
f(x) dx A = Area under the curve between coordinate-x a to b.
f(x) dx A = 2 f(x) dx
2 f(x) dx A = f(x) dx + f(x) dx
f(x) dx A = - f(x) dx
2. Given that f(x) dx = 4, find the value of
a) 3f(x) dx
= 3 f(x) dx
= 3
=
b) f(x) dx
= f(x) dx
=
c) f(x) dx + f(x) dx
= f(x) dx
=
13
a cb0
a b0
a b0
a b-a-b
y
x
a cb0
a b
y
a
b
x
x
x
y
y
0
0
x
y
x
y
xy
0
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
3. Shade the area of the region between the curve and x-axis.
a) b) c) d)
A = y dx A = y dx A
= - y dx + y dx
A
= y dx - y dx
Shade the area of the region between the curve and y-axis.
e) f) g) h)
A = x dy A = x dy A
= - x dy + x dy
A
= x dy - x dy
4. State the area , A ,of the shaded region in each of the following in definite integral form.
a) b) c)
A = f(x)dx - g(x)dx A= A =
5. Shade the area of the region between the curve and straight line.
a) b) c)
A = f(x)dx + g(x)dx A = f(x)dx + g(x)dx A = g(x)dx - f(x)dx
14
-a b a b a b-a b
00
c
-a
b
-a
b
0a
bc
0-a
b
c
0
-a b 0 b a b
y=f(x)y=g(x)
y=f(x)
y=g(x)
0
a b 0 c -a b
y=f(x)
y=g(x)
y=f(x)y=g(x)
00 -a b c
y=f(x)y=g(x)
x
y
x x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
x
y
x x x
y
y y
y=f(x)y=g(x)
0 0
0
y
y
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
6. Find the area of the shaded region.
a). y
x
Solution :
Area = y dx
= dx
= [ x + x ]
= [ ]
= [ ] - [ ]
= ( ) - ( )
= unit2
b). y
x
Solution :
Area = x dy
= dy
= [ y - 2 y ]
= [ ]
= [ ] - [ ]
= ( ) - ( )
= unit2
c). Find the area of the shaded region between the curve y = x2 – x , the x-axis and the coordinates x = 0 and x = 3.
y
x
Solution :
d). Find the area enclosed between the curve y = x2 + 2 and the line y = 4x -1.
y
x
Solution :
15
10 0
y = 4x -1y= x2 + 2
1 21
3
y = x2 + 1
y2 = x + 2
3 1 3
y2 = x + 2y2 - 2 = x x = y2 - 2
y = x2 + 1
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
Learning Outcome3.2.5 Determine the volumes of revolutions using formula.
Volume of revolution, y2 dx . Area under the graph, y dx .
16
Process ofIntegration,
definite integral , y dx
Process ofIntegration,
definite integral , y2 dx
a b
Applications of integration
ba
y
x
y
x
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
y y = f(x)
x
Equation, y = f(x)
Function/Equation, y
Gradient of tangent, m=
y
m =
x
Volume of revolution
17
Volume of revolution, V = y2 dx
where y = f(x) and V is the volume of solid generated when the curve y = f(x) between limits x = a and x = b is rotated completely round the x-axis.
Process ofDifferentiation
Process ofIntegration
x
y
a b
Volume of revolution, V = x2 dy
where y = f(x) and V is the volume of solid generated when the curve y = f(x) between limits y = a and y= b is rotated completely round the y-axis.
y = f(x)
0 x
y
a
b
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
1 a) Draw the solid of revolution when the shaded region is rotated through 360 about x-axis. i ii iii iv
1 b) Draw the solid of revolution when the shaded region is rotated through 360 about y-axis.i ii iii iv
2. Arrange and rewrite the process of integration to find the volume of revolution.
a) Find the volume of revolution generated when the part of the curve y = x2 from x =1 to x = 2 is rotated through 360 about x-axis.
1. V = V = y2 dx
2. = (x 2)2 dx
3. = [ ]
4. = (x 2 )2 dx
5. = x 4 dx
6. = [ ]
7. = [ ( ) - ( ) ]
8. = [ ]
9. = unit3 [ ( ) - ( ) ]
18
y
x0
y
x0
y
x0
y
x0
y
x0
y
x0
y
x0
y
x0
y
10 x2
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
3 Find the volume of revolution generated when the part of the curve y = x2 from y =1 to y = 2 is rotated through 360 about y-axis.
1. V = V = x2 dy
2. = [ ]
3. = (y ) dy
4. = [ ]
5. = [ ( ) - ( ) ]
6.= [ ]
7. = unit3 [ ( ) - ( ) ]
4 a) Find the volume generated when the shaded region is rotated about the x-axis through 360 ْ.
V = y2 dx
= ( )2 dx
= ( ) dx
= [ ]
= {[ ] – [ ( ] }
= { [ ] – [ ] }
= unit3
b) Find the volume generated when the shaded region is rotated about the y-axis through 360 ْ.
19
0
2
1
x
y
0 2x
yy = x2 + 1
y
rotated about the x-axis
rotated about the y-axis
WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration
V = x2 dy
= ( ) dy
= [ ]
= {[ ( )] – [( )] }
= { [ ] – [ ] }
= unit3
c) Find the volume generated when the shaded region is rotated about the y-axis through 360 ْ.
V = x2 dy
=
=
=
=
=
=
20
0
2
x
yy2 = x
0
2
x
y = x2
rotated about the y-axis