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Specific Objectives

By the end of the topic the learner should be able to:

(a) Carry out the process of differentiation;

(b) Interpret integration as a reverse process of differentiation;

(c) Relate integration notation to sum of areas of trapezia under a curve;

(d) Integrate a polynomial;

(e) Apply integration in finding the area under a curve,

(f) Apply integration in kinematics.

Content

(a) Differentiation

(b) Reverse differentiation

(c) Integration notation and sum of areas of trapezia

(d) Indefinite and definite integrals

(e) Area under a curve by integration

(f) Application in kinematics.

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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

 

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Integration reverses this process and we say that the integral of 2x is .

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

Example

Find y if is:

  1.  
  2.  

Solution

  1.  

    Then, y =

     

  2.  

    Then, y =

     

     

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 ,then, where c is a constant and n.since c can take any value we call it an arbitrary constant.

Example

Integrate the following expression

  1.  
  2.  
  3. 2x +4

Solution

 

Then, y =

=

=

B.)

Then, y =

=

= –

 

C.) 2x +4

Then, y =

=

=

Example

Find the equation of a line whose gradient function is and passes through (0,1)

Solution

Since ,the general equation is y =.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 =

Example

Find v in terms of h if and V =9 when h=1

Solution

The general solution is

V =

=

V= 9 when h= 1.Therefore

 

9 = 5 + c

4 = C

Hence the particular solution is

V

 

 

 

 

 

 

 

Definite and indefinite integrals

It deals with finding exact area.

Estimate the area shaded beneath the curve shown below

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The area is divided into rectangular strips as follows.

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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

 

The symbol

Thus means integrate the expression for y with respect to x.

 

 

The expression ,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

 

The following steps helps us to solve it

  1. Integrate with respect to x , giving
  2. Place the integral in square brackets and insert the limits, thus

 

  1. Substitute the limits

    X = 6 gives

    x = 6 gives

  2. Subtract the results of the lower limit from that of upper limit, that is;

    (162 + c) – (

We can summarize the steps in short form as follows:

=

=

=150

Example

  1. Find the indefinite integral
  2.  
  3.  

     

  4. Evaluate
    1.  
    2.  
    3.  

     

Solution

 

 

 

 

Evaluate

 

 

 

  1.  

    4 + 10 -4 ) – ( -)

     

     

  2.  

    = (27 – 18 +15) – (8 – 8 +10)

    = 14

 

 

 

 

Area under the curve

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

 

Solution

 

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2 4

 

The area is given by;

 

 

Example

Find the area of the region bounded by the curve , the x axis x = 1 and x = 2

Solution

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The area is given by;

 

= (4 – 8 + 4) – (

= 0 – =

Note;

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

Example

Find the area enclosed by the curve the x – axis and the lines x = 4 and x =10.

Solution

The required area is shaded below.

 

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Area =

 

 

 

 

Example

Find the area enclosed by the curve y and the line y =x.

Solution

The required area is

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To find the limits of integration, we must find the x co-ordinates of the points of intersection when;

 

 

 

 

The required area is found by subtracting area under y = x from area under y =

The required area =

 

 

 

 

 

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

S =

Substituting S = 4, t = 0 to get C;

 

4 = C

Therefore.

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 10 m/.

Solution

  1. = -10 ( since the ball is projected upwards)

    Therefore, v =-10 t + c

    When t = 0, v = 40 m/s

    Therefore, 40 = 0 + c

    40 = c

    1. The expression for velocity is v = 40 – 10t
    2. Since

       

      When t = 0 , S = 0

      C = 0

      The expression for displacement is

       

  2. Since v = 40 – 10t
    1. When t = 2

      v = 40 – 10 (2)

      = 40 – 20

      = 20 m/s

       

      S =40t –

      = 40 (2) – 5 (

      = 80 – 20

      = 60 m

    2. When t = 5

      V = 40 – 10 (5)

      = -10 m/s

      S

       

      = 75 m

    3. When t = 8

      V = 40

 

S

= 320 – 320

= 0

  1. 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.  

      S=4t + c

      Since S = 5 m when t =2;

      5 = 4 (2) + C

      5 – 8 = C

      -3 = C

      Thus, S =4t – 3

      1. I.)

       

       

      II.)




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EcoleBooks | Mathematics Form 1-4 : CHAPTER SIXTY FOUR - INTERGRATION

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