# Integration - Mathematics Form 4 Notes

## Introduction

• The process of finding functions from their gradient (derived) function is called integration
• Suppose we differentiate the function y=x2. We obtain dy/dx=2x
• Integration reverses this process and we say that the integral of 2x is x2.

• From differentiation we know that the gradient is not always a constant. For example, if dy/dx = 2x, then this comes from the function of the form y=x2 + c, Where c is a constant.

Example

Find y if dy/dx is:

1. 3x2
2. 4x3

Solution

1. dy3x2
dx
Then, y =x3 + c
2. dy4x3
dx
Then, y =x4 + c

Note;

• To integrate we reverse the rule for differentiation. In differentiation we multiply by the power of x and reduce the power by 1.
• In integration we increase the power of x by one and divide by the new power.
• If dy/dx = xn,then y = Xn+1 + c, where c is a constant and n≠ -1.since c can take any value we call it an arbitrary constant
n+1

Example

Integrate the following expression

1. 2x5
2. x-1
3. 5x3- 2x +4

Solution

1. dy =2x5
dx
Then, y = 2x5+1 + C
5+1
= 2x+ C
6
= x+ C
3
2. dyx-2
dx
Then, y =  x-2+1+ C
-2+1
=x-1+ C
-1
= -x-1 + C
3. dy = 5x- 2x +4
dx
Then, y =5x3+1 2x1+1 − 4x0+1+ C
3+1      1+1       0+1
=5/4x42/2x2 + 4x + C
=5/4x4 − x2 + 4x + C

Example

Find the equation of a line whose gradient function is dy/dx = 2x + 3 and passes through (0,1 )

Solution
Since dy/dx = 2x + 3,the general equation is y =x2 + 3x + C. The curve passes through ( 0,1 ). Substituting these values in the general equation ,we get 1 = 0 + 0 + C
1 = C
Hence, the particular equation is y =
x2 + 3x + 1

Example

Find v in terms of h if dV/dh =3h2 + 4 and V =9 when h=1

Solution

The general solution is
V =
3/3h3 + 4h + C
= h3 + 4h + C
V = 9 when h = 1 .Therefore
9 = 3 + 4 + C
9 = 5 + C
4 = C
Hence the particular solution is ;
V
= h3 + 4h + 1

## Definite and Indefinite Integrals

• It deals with finding exact area.
• Estimate the area shaded beneath the curve shown below

• The area is divided into rectangular strips as follows.

• The shaded area in the figure above shows an underestimated and an overestimated area under the curve.
• The actual area lies between the underestimated and overestimated area. The accuracy of the area can be improved by increasing the number of rectangular strips between x = a and x = b.
• The exact area beneath the curve between x = a and b is given by
bayδx
• The symbol ∫ is an instruction to intergrate.
• Thus ∫ydx means integrate the expression for y with respect to x.
• The expression bayδx, where a and b are limits, is called a definite integral. ‘a’ is called the lower limit while 'b' is the upper limit. Without limits, the expression is called an indefinite integral.

Example

62(2x2+3 )dx
The following steps helps us to solve it

1. Integrate 2x2 + 3 with respect to x , giving 2/3x3 + 3x + c.
2. Place the integral in square brackets and insert the limits, thus
[2/3x3+3 x+c]62
3. Substitute the limits ;
X = 6 gives 2/3 x 63 + 3 x 6 + c = 162 + C
x = 2 gives 2/3 x 23 + 3 x 2 + c = 34/+ C
4. Subtract the results of the lower limit from that of upper limit, that is;
(162 + C) – (34/3 + C ) = 1502/3

We can summarize the steps in short form as follows:

62(2x2+3 )dx = [2/3x3+3 x+c]62
=[2 x 63+(3 x 6)] – [2 x 23 +3 x 2]
3                           3
=150 2/3

Example

1. Find the indefinite integral
1. ∫(x2 + 1) dx
2. ∫( x+ 4x) dx
2. Evaluate
1. 10 ( x4  5) dx
2. 2-1(− x3+5x−2) dx
3. 32(3x24x+5) dx

Solution

1.
1. (x2 + 1) dx = x3+ x + c
3
2. ( x2 + 4x) dx = x4+ 2x2 + c
4
2.
1. [x5 − 5x]10
5
(1/−5) − (0/5 − 0)
= -44/5
2. x=[-x4+ 5x2 2x]2-1
4      2
(–4 + 10–4 ) – (–1/4 + 5/2+ 2)
= 2 – 41/4
= –21/4
3. 3x2 4x2+5x]32
3     2
= (27 – 18 +15) – (8 – 8 +10)
= 14

## Area Under the Curve

Find the exact area enclosed by the curve y = x2,the axis, the lines x = 2 and x = 4

Solution

The area is given by;
43x2dx=[1/3x3]42
64/3 - 8/3= 18 square units

Example

Find the area of the region bounded by the curve = x3 - 3x2 + 2x, the x axis x = 1 and x = 2

Solution

The area is given by;
21( x3  3x2 + 2x)dx = [x4x3+x2]21
4
= (4 – 8 + 4) – (¼- 1 + 1)
= 0 - ¼ =

Note;

The negative sign shows that the area is below the x – axis. We disregard the negative sign and give it as positive. The answer is ¼ square units.

Example

Find the area enclosed by the curve x2 - 10x + 9,the x – axis and the lines x = 4 and x =1 0.

Solution

The required area is shaded below.

Area = 94(x2-10x+9)dx+∫109(x2-10x+9)dx
=
[x3−5x2+9x]94 + [x35x2+9x]109
3                        3
= [(243−405+81)−(64/3−80+36)] + [(1000/3 − 500+90 ) − (243 − 405+81)]
= -581/3 + 41/3 (Drop negative sign for area under x - axis)
= 622/3 square units

Example

Find the area enclosed by the curve y = 9x−x2 , and the line y =x.

Solution

The required area is

To find the limits of integration, we must find the x co-ordinates of the points of intersection when;
x = 9x - x2
→0 = 8x - x2
0 = x (8 - x)
x = 0 or x = 8
The required area is found by subtracting area under y = x from area under y = = 9x - x2
The required area =80(9x-x2)dx −∫80xdx
[9x2- x3]80 −[x3]80
2     3          2
= 1171/3 − 32
= 851/3 square units

## Application in Kinematics

The derivative of displacement S with respect to time t gives velocity v, while the derivative of velocity with respect to time gives acceleration, a

Differentiation.            Integration

Displacement.                Displacement
↓↑
Velocity.                         Velocity.
↓↑
Acceleration.                  Acceleration.

Note;

Integration is the reverse of differentiation. If we integrate velocity with respect to time we get displacement while if velocity with respect to time we get acceleration.

Example

A particle moves in a straight line through a fixed point O with velocity ( 4 – 1 )m/s.Find an expression for its displacement S from this point, given that S = when t = 0.
Solution
Since dS/dt = 4 - t
S = 4t - t2 + C
2

Substituting S = 4, t = 0 to get C;
4 = 4 x 0 - 02 + C
2
4 = C
Therefore
S = 4t - t2 + 4.
2
Example

A ball is thrown upwards with a velocity of 40 m s

1. Determine an expression in terms of t for
1. Its velocity
2. Its height above the point of projection
2. Find the velocity and height after:
1. 2 seconds
2. 5 seconds
3. 8 seconds
3. Find the maximum height attained by the ball. (Take acceleration due to gravity to be 1 0 m/s2.

Solution

1. dv/dt = -10 (since the ball is projected upwards)
dt
Therefore, v =-10t + C
When t = 0, v = 40 m/s
Therefore, 40 = 0 + C
40 = C
The expression for velocity is v = 40 – 1 0t
Since
dS/dt = v = 40 - 10t;
S = 40t - 5t3 + c
When t = 0 , S = 0 C = 0
The expression for displacement is ;
S = 40t - 5t2
2. Since v = 40 – 1 0t
1. When t = 2
v = 40 – 10(2)
= 40 – 20
= 20 m/s
S = 40t - 5t2
= 40(2) – 5(2)2
= 80 – 20
= 60 m
2. When t = 5
V = 40 – 10 (5)
= -10 m/s
S = 40 (5) - 4 (5)2
= 200 - 125
= 75 m
3. When t = 8
V = 40 – 10(8)
= -40 m/s
S= 40(8) - 5 (8)2
= 320 – 320
= 0
3. Maximum height is attained when v = 0.
Thus , 40 – 10t = 0
t= 4
Maximum height S = 160 – 80
= 80 m

Example

The velocity v of a particle is 4 m/s. Given that S = 5 when t =2 seconds:

1. Find the expression of the displacement in terms of time.
2. Find the :
1. Distance moved by the particle during the fifth second.
2. Distance moved by the particle between t =1 and t =3.

Solution

1. dS/dt = 4t + c
S=4t + c
Since S = 5 m when t =2;
5 = 4 (2) + C
5 – 8 = C
-3 = C
Thus, S =4t – 3
2.
1. [4t-3] 5 4 = [(20-3 ) - (16-3 )]
= 17 - 13
= 4 m
2. [4t-3] 3 1 = [(12-3 ) - (4-3 )]
= 9 - 1

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