Integration - Mathematics Form 4 Notes

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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.
    introduction integration
  • 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
    definite integral
  • The area is divided into rectangular strips as follows.
    definite integral 2
  • 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

area under curve 1

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

area under curve 2

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 under curve 3
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

area under curve 4

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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