Integration

CHAPTER 3: INTEGRATION 3.0 Basic Concepts

Integration is the reverse process of differentiation

1. Given , therefore ∫ 2. Given , therefore . 1. Given , Find . Solution : 2. Given , Find . 3. Given , Find . 4. Given , Find . 5. Given , Find . 6. Given , Find . 7. 7. Given , Find . 8. Given , Find .

9. Given , Find . 10. Given , Find . 3.1. Indefinite Integral 3.1.2 (A) Integration of

Exercise : 1. Determine the following integrals ;

1. = = 2. 3. 4. 5. = = 6. 7. 8.

3.1.2 (B) Integration of

Exercise : 2. Determine each of the following integrals ;

1. = 2. 3. 4. 5. = = = = 6. 7. 8. 9. = = = = 10. 11. 12.

3.1.3. Integration of Algebraic Expression

Application : Basic techniques of integration on each term. Expand and simplify the given expression where necessary.

e.g.

=

1. 2. 3. 4.

5.

9. 10.

11. 12.

3.1.4 (A) Determine the Constant of Integration

e.g. 1. Given that

Find y in terms of x. Solution ; hence 2. Given that Find y in terms of x. 3. Given that Find y in terms of x.

4. Given that Find y in terms of x. 5. Given that Find y in terms of x. 6. Given that Find y in terms of x. 7. Given that Find p in terms of v. 8. Given that Find y in terms of x. 9. Given that

Find in terms of x if y=12 when x=5

e.g. 1. Given

Find the value of y when x=3 Solution ;

2. Given

Find the value of y when x= -1

-3

3. Given

Find the value of y when x= 0

3

4. Given

Find the value of y when x= 2

5. Given

Find the value of y when x= 3

6. Given

Find the value of y when x= -1

-7 25 8/3

3.1.5. Determine the equation of curve from gradient function

e.g. 1. The gradient function of a curve which passes through A(1,5) is . Find the equation of the curve.

Solution :

2. The gradient function of a curve which passes through B(1,-3) is . Find the equation of the curve.

3. The gradient function of a curve which passes through P(2,3) is . Find the equation of the curve.

4. The gradient function of a curve which passes through Q(1,6) is . Find the equation of the curve.

5. The gradient function of a curve which passes through R(-1,4) is . Find the equation of the curve.

6. The gradient function of a curve which

passes through S(1,10) is . Find the equation of the curve.

7. The gradient function of a curve which

passes through X(-1,7) is . Find the equation of the curve.

8. The gradient function of a curve is

. Find the equation of the curve if it passes through the point (-2,3)

3.1.6. Integration of expressions of the form

Using formula :

e.g. 1. 2. 3.

4.

+c

=

5.

6.

e.g. 1. 2. 52 3. 64 4. 20 5. 3/2 6

.

7. 22.5 8. -10 9 25/3

3.2.1 Definite Integral of the form

e.g. 1.

203/4 24 4. 1280 5. 5/12 6. -65/3 7

.

.

.

1 5 3.75

3.2.2(A) Application of Definite Integral

Exercise : 1. Given

-6 12 8 d. 14 e. 7 f. 44 Exercise : 2. Given

a. 20 b. -4 c. 8 d. 0 e. 6 f. 16

Exercise : 3. Given

Find the value of ;

a.

10

b. 3

c. 2

5

d.

15/2

3.2.2 (B) Applications of Definite Integral

1. Given that

Where k >-1 , find the value of k . Solution ;

2. Given that

4

3. Given that

Where k >0 , find the value of k .

4

4. Given that

Where k >0 , find the value of k .

5. Given that

find the value of k . Solution ;

; k = 1/4

6. Given that

find the value of k

2

7. Given that

find the value of k .

8. Given that

8/21

-2

3.3 Area under a curve using Definite Integral

3.3.1 Area under a curve bounded by x-axis

Exercise : 1. Find the area of the shaded region.

Note : 1. Area above the x-axis has a positive value. 2. Area beneath/below the x-axis has a negative value.

y

x

a) Solution: = b) 20 c) d) y x 0 1 3 x y -3 2 y x 0 1 5 y x 0 5

d) e) y x 0 -2 y x 0 4

3.3.2 Area under a curve bounded by y-axis

Note : 1. Area on the right side of the y-axis has a positive value.

2. Area on the left side of the y-axis has a negative value. y x 0 a b

Solution: a) 10 b) 6 d) y x 0 2 y x 0 2 -1 y x 0 2 -2 y x 2 -2

e) f) 0 y x 1 -1 x 0 9 4 y

3.3.3. Area between two graphs

Exercise : 3. Find the area of the shaded region a) b) y x 0 a b y x 3 0 y x 0 y x 0 -1 1

c) d) 3.4 Volume of Revolution

A. Region bounded by a curve when is rotated completely about x-axis. y x 0 y x 0 a b y=f(x)

Exercise 1 : Find the volume of solid generated when the shaded region revolves 360o _about

x-axis.

a) b)

c)

Volume of shaded region =

y x 0 2 y x 0 3 6 y x 0 -1 3 1 3 y x 0

B. Region bounded by a curve when is rotated completely about y-axis

Exercise 2 : Find the volume of solid generated when the shaded region revolves 360o _about

y-axis. a) b) c) d) a y x 0 b

Volume of shaded region =

y x 0 3 y x 0 1 1 y x 0 y x 0 1 -4

Extra exercises : 1)

The diagram shows part of the curve y = 4–x2

and a straight line y + 2x = 6. Find the

volume generated when the shaded region is rotated through 360o _{about x-axis.}

Solution : y + 2x = 6 y = 6 – 2x Volume of a cone 2) 6 y x -2 2 3 4 y x 2 1

Find the volume generated when the shaded region is rotated through 360o _{about the}

x-axis.

3)

Find the volume of solid generated when the shaded region is revolve 360o _{about x-axis.}

4) y x 1 =36 y x 0

Find the volume of solid generated when the shaded region is revolved 360o _{about y-axis.}

5)

Find the volume of solid generated when the shaded region is revolved 360o _{about y-axis.}

6) y x 2 -2 y x 0

Find the volume of solid generated when the shaded region is revolved 360o _{about y-axis.}