Area Approximation - Mathematics Form 4 Notes

• Introduction
• Approximating Area by Trapezium Method.
• The Mid - Ordinate Rule
• Past KCSE Questions on the Topic.

Introduction

- Estimation of areas of irregular shapes such as lakes, oceans etc. using counting method. The following steps are followed

• Copy the outline of the region to be measured on a tracing paper
• Put the tracing on a one centimeter square grid shown below
  
• Count all the whole squares fully enclosed within the region
• Count all the partially enclosed squares and take them as half square centimeter each
• Divide the number of half squares by two and add it to the number of full squares.
  Number of compete squares = 4
  Number of half squares = ¹⁶/₂ = 8
  Therefore the total number of squares = 25 + 8 = 33
  The area of the land mass on the paper is therefore 33 cm²
  
  Note;
• The smaller the subdivisions, the greater the accuracy in approximating area.

Approximating Area by Trapezium Method.

- Find the area of the region shown, the region may be divided into six trapezia of uniform as shown

- The area of the region is approximately equal to the sum of the areas of the six trapezia.

Note;
- The width of each trapezium is 2 cm, and 4 and 3.5 are the lengths of the parallel sides of the first trapezium.

The area of the trapezium A = ½ x 2(4+3.5) = 7.5 cm²
Area of the trapezium B = ½ x 2(3.5+3.5)  = 6.7 cm²
Area of the trapezium C = ½ x 2(3+1.5) = 6.2cm²
Area of the trapezium D = ½ x 2(3+1.5) = 4.5 cm²
Area of the trapezium E = ½ x 2(1.5+0.8) = 2.3 cm²
Area of the trapezium F = ½ x 2(0.8+3.5) = 1.3 cm²
Therefore, the total area of the region is
(7.5 +6.7 +6.2 +4.5 +2.3 +1.3)cm² = 28.5 cm²

-  If the lengths of the parallel sides of the trapezia (ordinates) are y₁, y₂ ,y₃ ,y₄ ,y₅, y₆ y₇

Note;
- In trapezium rule, except for the first and last lengths, each of the other lengths is counted twice. Therefore, the expression for the area can be simplified to:
½ x 2 {(y₁+ y₇)+2 (y₂+ y₃+ y₄+ y₅+ y₆)}

In general, the approximate area of a region using trapezium method is given by:
A =½h{(y₀+ y_n)+2(y₁+ y₂+ y₃+…………. y_n-1 )}
Where h is the uniform width of each trapezium,
y₀ - y_n are the first and last length respectively.

- This method of approximating areas of irregular shape is called trapezium rule.

Example

A car start from rest and its velocity is measured every second from 6 seconds. 

Use the trapezium rule to calculate distance travelled between t = 1 and t = 6

Note;
- The area under velocity – time graph represents the distance covered between the given times.
- To find the required displacement, we find the area of the region bounded by graph, t =1 and t =6

Solution

Divide the required area into five trapezia, each of with 1 unit. Using the trapezium rule;
A = ½ h{(y₁+ y₆)+2 (y₂+ y₃+ y₄+y₅ )}

The required displacement = ½h{(12+47 ) + 2( 24+35+41+45 )}
= ½( 59 + 2 x145 )
= 174.5 m

Example

Estimate the area bounded by the curve y = ½x² + 5, the x – axis, the line x =1 and x = 5 using the trapezium rule.

Solution

To plot the graph  y = ½x2 + 5, make a table of values of x and the corresponding values of y as follows:

By taking the width of each trapezium to be 1 unit, we get 4 trapezia A, B , C and D .The area under curve is approximately;

A = ½h{(y1+ y5)+2(y2 + y3+ y4 )}
=  ½(5.5+ 17.5)+ 2(7 + 9.5 + 13) sq.units
= 1 sq.units
½(23.0+59)
= 41 sq.units

The Mid - Ordinate Rule

- The area OPQR is estimated:

- The area of OPQR is estimated as follows

• Divide the base OR into a number of strips, each of their width should be the same .In the example we have 5 strips where
  h = length of the base OR 
        number of strips
• From the midpoints of OE ,EF ,FG ,GH and HR , draw vertical lines ( mid- ordinates) to meet the curve PQ as shown above
• Label the mid-ordinates y₁,y₂,y₃,y₄ and y₅
• We take the area of each trapezium to be equal to area of a rectangle whose width is the length of interval (h) and the length is the value of mid –ordinates. Therefore, the area of the region OPQR is given by;
  
  A = (y₁×h) + (y₂ ×h) + (y₃×h) + (y₄ ×h) + (y₅ ×h)
  = h (y₁ + y₂ + y₃ + y₄ + y₅ )

- This the mid –ordinate rule h(y₁ + y₂ + y₃ + y₄ + y₅).

Note:

The mid-ordinate rule for approximating areas of irregular shapes is given by ;

Area = (width of interval) x (sum of mid – ordinates)

Example

Estimate the area of a semi-circle of radius 4 cm using the mid – ordinate rule with four equal strips, each of width 2 cm.

Solution

The above shows a semicircle of radius 4 cm divided into 4 equal strips, each of width 2 cm. The dotted lines are the mid-ordinates whose length are measured.
By mid- ordinate rule;
= h (y₁ + y₂ + y₃ + y₄ + y₅ )
= 2 (2.6 + 3.9 + 3.9 + 2.6)
= 2 x 13
= 26 cm²
The actual area is πr² = 3.142 ×4²
                                           2
= 25.14 cm² to 4 s.f

Example  

Estimate the area enclosed by the curve y =½ x² + 1 , x = 0, x = 3 and the x – axis using the mid-ordinate rule.

Solution

Take 3 strips. The dotted lines are the mid – ordinate and the width of each of the 3 strips is 1 unit.

By calculation, y₁ ,y₂ and y₃ are obtained from the equation;
y =½ x² + 1

When x = 0.5, y₁ = ¹/₂ x(0.5)² + 1
= 1.125
When x = 1.5, y₁ = ¹/₂ x(1.5)² + 1
= 2.125
When x = 2.5, y₁ = ¹/₂ x(0.5)² + 1
= 4.125

Using the mid ordinate rule the area required is
A = 1 (y₁ + y₂ + y₃)
= 1(1 .125 + 2.125 + 4.125)
= 7.375 square units

Past KCSE Questions on the Topic.

1. The shaded region below represents a forest. The region has been drawn to scale where 1 cm represents 5 km. Use the mid – ordinate rule with six strips to estimate the area of forest in hectares. (4 marks)
   
   
2. Find the area bounded by the curve y=2x³ – 5, the x-axis and the lines x=2 and x=4.
3. Complete the table below for the function y=3x² – 8x + 10 (1 mk)
   

   x	0	2	4	6	8	10
   y	10	6		70		230

   Using the values in the table and the trapezoidal rule, estimate the area bounded by the curve y= 3x²– 8x + 1 0 and the lines y=0, x=0 and x=104.
4. Use the trapezoidal rule with intervals of 1 cm to estimate the area of the shaded region below
   
   
5.  
   
   1. Find the value of x at which the curve y= x – 2x² – 3 crosses the x - axis
   2. Find ∫(x² – 2x – 3)dx
   3. Find the area bounded by the curve y = x² – 2x – 3, the axis and the lines x= 2 and x = 4.
6. The graph below consists of a non- quadratic part (0 ≤ x ≤ 2) and a quadrant part (2 ≤ x 8). The quadratic part is y = x² – 3x + 5, 2 ≤ x ≤ 8
   
   
   1. Complete the table below
      

      x	2	3	4	5	6	7	8
      y	3

      (1 mk)
   2. Use the trapezoidal rule with six strips to estimate the area enclosed by thecurve, x = axis and the line x = 2 and x = 8 (3mks)
   3. Find the exact area of the region given in (b) (3mks)
   4. If the trapezoidal rule is used to estimate the area under the curve between x = 0 and x = 2, state whether it would give an under- estimate or an over- estimate. Give a reason for your answer.
7. Find the equation of the gradient to the curve Y= (x² + 1 ) (x – 2) when x = 2
8. The distance from a fixed point of a particular in motion at any time t seconds is given by
   S = t³ – 5t² + 2t + 5
           2t²
   Find its:
   1. Acceleration after 1 second
   2. Velocity when acceleration is Zero
9. The curve of the equation y = 2x + 3x², has x = -²/₃ and x = 0 and x intercepts. The area bounded by the axis x = -²/₃ and x = 2 is shown by the sketch below.
   
   Find:
   1. (2x + 3x²) dx
   2. The area bounded by the curve x – axis, x = - ²/₃ and x =2
10. A particle is projected from the origin. Its speed was recorded as shown in the table below
    

    Time (sec)	0	5	1 0	1 5	20	25	39	35
    Speed (m/s)	0	2.1	5.3	5.1	6.8	6.7	4.7	2.6

    Use the trapezoidal rule to estimate the distance covered by the particle within the 35 seconds.
11. 1. The gradient function of a curve is given by ^dy/_dx = 2x² – 5. Find the equation of the curve, given that y = 3, when x = 2
    2. The velocity, vm/s of a moving particle after seconds is given: v = 2t³ + t² – 1 . Find the distance covered by the particle in the interval 1 ≤ t ≤ 3
12. Given the curve y = 2x³ + ¹/₂x² – 4x + 1 . Find the:
    1. Gradient of curve at {1 , -¹/₂}
    2. Equation of the tangent to the curve at {1, - ¹/₂}
13. The diagram below shows a straight line intersecting the curve y = (x-1)² + 4 At the points P and Q. The line also cuts x-axis at (7, 0) and y axis at (0, 7)
    
    1. Find the equation of the straight line in the form y = mx +c.
    2. Find the coordinates of p and Q.
    3. Calculate the area of the shaded region.
14. The acceleration, a ms^-2, of a particle is given by a =25 – 9t², where t in seconds after the particle passes fixed point O. If the particle passes O, with velocity of 4 ms^-1, find
    1. An expression of velocity V, in terms of t
    2. The velocity of the particle when t = 2 seconds
15. A curve is represented by the function y = ¹/₃ x³ + x² – 3x + 2
    1. Find ^dy/_dx
    2. Determine the values of y at the turning points of the curve y = ¹/₃x³ + x² – 3x + 2
    3. In the space provided below, sketch the curve of y = ¹/₃ x³ + x² – 3x + 2
16. A circle centre O, ha the equation x2 + y2 = 4. The area of the circle in the first quadrant is divided into 5 vertical strips of width 0.4 cm
    1. Use the equation of the circle to complete the table below for values of y
       correct to 2 decimal places
       

       X	0	0.4	0.8	1 .2	1 .6	2.0
       Y	2.00			1.60		0

    2. Use the trapezium rule to estimate the area of the circle
17. A particle moves along straight line such that its displacement S metres from a given point is S = t³ – 5t² + 4 where t is time in seconds
    Find
    1. The displacement of particle at t = 5
    2. The velocity of the particle when t = 5
    3. The values of t when the particle is momentarily at rest
    4. The acceleration of the particle when t = 2
18. The diagram below shows a sketch of the line y = 3x and the curve y = 4 – x² intersecting at points P and Q.
    
    1. Find the coordinates of P and Q
    2. Given that QN is perpendicular to the x- axis at N, calculate
       1. The area bounded by the curve y = 4 – x², the x- axis and the line QN (2 marks)
       2. The area of the shaded region that lies below the x- axis
       3. The area of the region enclosed by the curve y = 4 – x², the line y – 3x and the y-axis.
19. The gradient of the tangent to the curve y = ax³ + bx at the point (1 , 1 ) is -5. Calculate the values of a and b. 2007
20. The diagram on the grid below represents as extract of a survey map showing two adjacent plots belonging to Kazungu and Ndoe. The two dispute the common boundary with each claiming boundary along different smooth curves coordinates (x, y₁) and (x, y₂) in the table below, represents points on the boundaries as claimed by Kazungu Ndoe respectively.
    

    X	0	1	2	3	4	5	6	7	8	9
    Y1	0	4	5.7	6.9	8	9	9.8	10.6	11.3	12
    Y2	0	0.2	0.6	1.3	2.4	3.7	5.3	7.3	9.5	12

    1. On the grid provided above draw and label the boundaries as claimed by Kazungu and Ndoe.
    2.  
       1. Use the trapezium rule with 9 strips to estimate the area of the section of the land in dispute
       2. Express the area found in b (i) above, in hectares, given that 1 unit on each axis represents 20 metres
21. The gradient function of a curve is given by the expression 2x + 1 . If the curve passes through the point (-4, 6);
    1. Find
       1. The equation of the curve
       2. The vales of x, at which the curve cuts the x- axis
    2. Determine the area enclosed by the curve and the x- axis
22. A particle moves in a straight line through a point P. Its velocity v m/s is given by v= 2 – t, where t is time in seconds, after passing P. The distance s of the particle from P when t = 2 is 5 metres. Find the expression for s in terms of t.
23. Find the area bonded by the curve y=2x – 5 the x-axis and the lines x=2 and x = 4.
24. Complete the table below for the function
    Y = 3x² – 8 x + 1 0
    

    X	0	2	4	6	8	1 0
    Y	1 0	6	-	70	-	230

    Using the values in the table and the trapezoidal rule, estimate the area bounded by the curve y = 3x² – 8x + 1 0 and the lines y – 0, x = 0 and x = 1 0
25.  
    
    1. Find the values of x which the curve y = x² – 2x – 3 crosses the axis
    2. Find (x² – 2 x – 3) dx
    3. Find the area bounded by the curve Y = x2 – 2x – 3. The x – axis and the lines x = 2 and x = 4
26. Find the equation of the tangent to the curve y = (x + 1 ) (x - 2) when x = 2
27. The distance from a fixed point of a particle in motion at any time t seconds is given by s = t – ⁵/₂t² + 2t + s metres
    Find its
    1. Acceleration after t seconds
    2. Velocity when acceleration is zero
28. The curve of the equation y = 2x + 3x², has x = - ²/₃ and x = 0, as x intercepts. The area bounded by the curve, x – axis, x = -²/₃ and x = 2 is shown by the sketch below.
    
    
    1. Find ∫(2x + 3x²) dx
    2. The area bounded by the curve, x axis x = -²/₃ and x = 2
29. A curve is given by the equation y = 5x³ – 7x² + 3x + 2
    Find the
    1. Gradient of the curve at x = 1
    2. Equation of the tangent to the curve at the point (1 , 3)
30. The displacement x metres of a particle after t seconds is given by x = t² – 2t + 6, t> 0
    1. Calculate the velocity of the particle in m/s when t = 2s
    2. When the velocity of the particle is zero,
       Calculate its
       1. Displacement
       2. Acceleration
31. The displacement s metres of a particle moving along a straight line after t seconds is given by s =3t + ³/₂t² – 2t³
    1. Find its initial acceleration
    2. Calculate
       1. The time when the particle was momentarily at rest.
       2. Its displacement by the time it comes to rest momentarily when t = 1 second, s = 1½ metres when t = ½ seconds
    3. Calculate the maximum speed attained