AM_F5_C3 (T).doc

WAJA 2009

ADDITIONAL MATHEMATICS

FORM FIVE

( Teacher’s Copy )

Name: ___________________________

Class : ___________________________

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 5) Chapter 3: Integration

3.1 Indefinite Integral

Learning Objective 3.1.0 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

You have probably wondered if it is possible to find a function given its gradient

function. For example, if the gradient function, = 6x + 2, is it possible to find y ?

Process of

differentiation

Process of

integration

must both being written. And read as the integral of 6x + 2 with respect to x.


2. Fill in the blanks in the following.

Function / Equation, y = f(x) 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 function.

(4x

+ c) = 4(10x ) dx f(x) = 4x + c

(5x2

+ c) = 10x

(6x -2) dx

f(x) = 5x2 + c

(3x2

– 2x + c) = 6x – 2

4 dx

f(x) = 4x 2 + 3x + c

(4x 2

+ 3x + c) = 8x + 3(8x + 3) dx f(x) = 3x2 – 2x + c

3

6x

6x(3x2+ 5)

12x3

12x3(3x4+ 4)

3

12x3 3x4 c


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 integral of the following.a) 3 x4 dx

Solution:

=> x4 dx = 3 ( ) + c

= 3 x 5 + c 5

b) dx

Solution:

=> x -4 dx = 2 x -4+1 + c -4 + 1 = 2 x -3 + c -3

= - x -3 + c

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 +

4

n dx = a ( ) + c , where c is a constant and n -1.

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

3 + 1

4

4

-2 + 1

-1

-2 + 1

-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.

c


d. dx

= x + c

= x + c

= x +

e. dx

= x + c

= x + c

= x +

f. dx

= 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 indefinite integral 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

= x2 dx + 3x dx - 2x -2 dx

= + 3( ) – 2( ) + c

= + 3( ) – 2( ) + c

= + 3( ) + 2x -1 + c

2. Arrange and rewrite the process of integration with respect to x.

5

dx = dx dx

5/2

5/2

5/2

3/2 + 1

2/5 + 1

7/5

7/5

7/5

3/2 + 1

1/2

1/2

2

-1/2 + 1

1/2

2/5 + 1

-1/2 + 1cc

c


Process of integration

1.

2. dx + dx

3.6 x (1 + 1) + 2 x + c (1 + 1)

4.3x2 + 2x + c

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.

6

(6x+2) dx

3x2 + 2x + c

dx + dx

6 x (1 + 1) + 2 x + c (1 + 1)

3 + 1 1 + 1 0 + 1

3 + 1 1 + 1 0 + 1

4 22

12

2 32 1

a) (4x3 + 5x2 – 2) dx

= 4x3 dx + 5x2 dx - 2 dx

= 4x 3 + 1 + 5x 2 + 1 - 2x + c 3+1 2+1

= 4( ) + x 3 - 2x + c

= x 4 + x 3 - 2x + c

b) (4x3 + 5x2 – ) dx

= 4x 3 + 1 + 5x 2 + 1 – 2x -2 + 1 + c 3 + 1 2 + 1 -2 + 1

= 4x 4 + 5x 3 - 2x -1 + c 4 3 -1

= x4 + x3 + 2x-1 + c

c

c


Function/Equation, y = 3x2 + 2x + c

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) = 3(1)2 + 2(1) + c

3 = 3 + 2 + c 3 - 5 = c

c = -2

2. Find the value of c.

Equation , y x y c

P(0,5) 6x + 2 y= 3x2 + 2x + c 0 5 5

Q(-1,6) 6x + 2 y= 3x2 + 2x + c -1 6 5

R(0,-3) 6x + 2 y= 3x2 + 2x + c 0 -3 -3

S(-1,-2) 6x + 2 y= 3x2 + 2x + c -1 -2 -3

Solution :

7

Process ofDifferentiation

Process ofIntegration

c is the y-intercept of the curve

Substitute the value y = 3 and x = 1.

y

PQ(-1,6)

5

x

RS(-1,-2) -3


=> = 3x2 + 2x + c

=> y = 3x2 + 2x + c

=> At point P(0,5), x = 0 , y = 5. 5 = 3(0)2 + 2(0) + c 5 = c

c = 5

=> At point Q(-1,6), x = -1 , y = 6.

6 = 3(-1)2 + 2(-1) + c 6 = 3 -2 + c 5 = c

c = 5

=> At point R(0,-3), x = 0 , y = -3.

-3 = 3(0)2 + 2(0) + c -3 = c

c = -3

=> At point P(-1,-2), x = -1 , y = -2.

-2 = 3(-1)2 + 2(-1) + c -2 = 3 -2 + c -3 = c

c = -3

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, (0) = 4(1)2 + (1) + c

0 = 4 + 1 + c - 5 = c

c = -5

The equation of the curve, y = 4x 2 + x - 5

2. Determine the equation of the curve that has the gradient function and passes through

the point P.

a) b)

8

Substitute the value y = 0 and x = 1.


The curve passing through Gradient function ,

The curve passing through Gradient function ,

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 = 6

Equation of the curve, y = + 3 x + 6

=> = 2x - 1

y = dx y = + c

Since the curve passing through P(-2 , 5), substitute the value y = 5 and x = -2.

= + c

5 = 4 + 2 + c

c = -1

Equation of the curve, y = x2 – x -1

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 (3)(5)

= (3x – 2 ) 5 + c 15

b) (4x + 5 dx

= (4x + 5 + c

(4)

= (4x + 5 + c

(4) ( )

= (4x + 5 + c 10

2. Fill in the blank.

a) (5x – 1)2 dx b) (4x + 5)3 dx c) (3x – 7) dx

9

n dx = (ax + b) n+1 + c , where c is a constant and n -1. (n + 1) (a)

x + 3

+ 3 x

6 + 3(0)

2x - 1

2( ) - x

5 (-2)2 – (-2)


= (5x - 1) +

= (4x + 5) +

= (3x – 7) ( + 1) +

= (5x - 1) + = (4x + 5) +

= (3x – 7) +

= (5x - 1) +

= (4x + 5) +

= (3x – 7) +

3. Match the indefinite integral 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 ]

= [ - + (2) ] - [ - + 0 ]

10

2 + 1

2 + 1

35

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

3+13+1

4

4

4

4

4

16

3

5/2

3/2 +1

3

c

c

c

c

c

c

c

c

c


= = [ - 6 + 2] - [ 0 ]

= -

2. Evaluate the definite integral.

(6x + 2) dx

1. [ x2 + 2x ]

2. [ 3x2 + 2x ]

3.[ 3(3)2 + 2(3) ] – [3(1)2 + 2(1) ]

4.[ 33 ] – [ 5 ]

5. 28

3. Evalute dx . Draw 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

[ x2 + 2x ]

28

[ 3x2 + 2x ]

[ 33 ] – [ 5 ]

[ 3(3)2 + 2(3) ] – [3(1)2 + 2(1) ]

dx

= 18

=[ 2 x2 = 2(3)2 – 2(0)2

= [ 4()


y y = f(x)

x

Equation, y = f(x)

Function/Equation, y


y

m =

x

Further notes on areas

12



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


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 ( 4 )

= 12

b) f(x) dx

= f(x) dx

= - 4

c) f(x) dx + f(x) dx

= f(x) dx

= 4

13

a cb0

a b0

a b0

a b-a-b

y

x

-6

3

6

3

a cb0x

ya b

y x

y

0

a

b

x0

y

x

y

x

y x


3. Shade the area of the region between the curve and the 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 the y-axis.

e) f) g) h)

A = x dy A = x dy A

= - xdy + x dy

A

= x dy - xdy

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 = f(x)dx - g(x) dx A = f(x)dx - g(x) dx

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

bc

0

-a b 0 b 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)

y

x

y

x

y

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)


6. Find the area of the shaded region.

a). y

x

Solution :

Area = y dx

= dx

= [ x + 1x ]

= [ + x ]

= [ + (2)] - [ + (1)]

= ( ) - ( )

= unit2

b).

y

x

Solution :

Area = x dy

= dy

= [ y - 2 y ]

= [ - 2y ]

= [ - 2 (3) ] - [ - 2 (1) ]

= ( 3 ) - ( )

= 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

Answer : unit2

d). Find the area enclosed between the curve y = x2 + 2 and the line y = 4x -1.

Answer : unit2

15

10 0

y = 4x -1y= x2 + 2

1 2 1

3

y = x2 +1

y2 = x + 2

3

x2 +1

3 2

13

2

1

2 3

12

y2 - 2

1 3

3

1

y2 = x + 2y2 - 2 = x x = y2 - 2

y = x2 + 1


Learning Outcome3.2.4 Determine the volumes of revolutions using formula.

Volume of revolution, y2 dx . Area under the graph, y dx .

16

a b

Applications of integration

ba

y

x

y

x


y y = f(x)

x

Equation, y = f(x)

Function/Equation, y


y

m =

x

Volume of revolution

17


definite integral , y dx


definite integral , y2 dx



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.

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


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 = y2 dx V = y2 dx

2.= (x 2 )2 dx (x 2)2 dx

3.= (x 2 )2 dx [ ]

4.= (x 4) dx (x 2 )2 dx

5. = [ ] x 4 dx

6.= [ ] [ ]

18

y

x0

y

x0

y

x0

y

x0

y

x0

y

x0

y

x0

y

x0

1

y

0 x2


7. = [ ( ) - ( ) ] [ ( ) - ( ) ]

8. = [ ( ) - ( ) ] [ ]

9. = [ ] unit 3 [ ( ) - ( ) ]

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 = x2 dy V = x2 dy

2. (y ) dy [ ]

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

= (x2 + 1)2 dx

= (x4 + 2x2 + 1) dx

= [ + 2 + x ]

= {[ ( )+ 2( ) + (2) ] – [ ( )+ 2( ) + (0) ] }

= { [ ] – [ 0 ] }

= unit3

b) Find the volume generated when the shaded region is rotated about the y-axis through 360 ْ.

19

0

2

1x

y

0 2x

yy = x2 + 1

y

rotated about the x-axis

rotated about the y-axis


V = x2 dy

= (y) dy

= [ ]

= {[ ( )] – [( )] }

= { [2] – [ 0 ] }

= 2 unit3

c) Find the volume generated when the shaded region is rotated about the y-axis through 360 ْ.

V = x2 dy

= (y2)2 dy

= (y4 ) dy

= [ ]

= {[ ( )] – [( )] }

= { [ ] – [ 0 ] }

= unit3

20

0

2

x

yy2 = x

0

2

x

y = x2

rotated about the y-axis

AM_F5_C3 (T).doc

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