Trapezoidal Rule

To find the area of the trapezium as shown in the figure, the base AD is divided into equal intervals of width S. The ordinates a,b,c,d,e,f,g are accurately measured. The approximation used in this rule is to assume that each strip is equal to the area of a trapezium. Therefore:

Area of a trapezium  = \frac{1}{2} (sum of parallel sides)  \times (perpendicular distance between the parallel sides)


trapezoidal-rule

Hence, the first strip's approximate area is \frac{1}{2}\left( {a + b} \right)S

For the second strip, the approximate area is \frac{1}{2}\left( {c + d} \right)S, and so on.

Therefore, the approximate area of ABCD

 = \frac{1}{2}\left( {a + b} \right)S + \frac{1}{2}\left( {b + c} \right)S + \frac{1}{2}\left( {c + d} \right)S + \frac{1}{2}\left( {d + e} \right)S + \frac{1}{2}\left( {e + f} \right)S + \frac{1}{2}\left( {f + g} \right)S

 = S\left( {\frac{{a + b}}{2} + \frac{{b + c}}{2} + \frac{{c + d}}{2} + \frac{{d + e}}{2} + \frac{{e + f}}{2} + \frac{{f + g}}{2}} \right)

 = S\left( {\frac{a}{2} + b + c + d + e + f + \frac{g}{2}} \right) = S\left( {\frac{{a + g}}{2} + b + c + d + e + f} \right)

Area  = (width of interval) [sum of first and last ordinate /2 + sum of remaining ordinates]

 

Example:

Find the area of a cross-section of a river along a line where the depths at equal intervals of 10m are noted as 0,7,11,15,5,0m, respectively.

 

Solution:

Width of each strip, S = 10m
Ordinates are  = 0,7,11,15,5,0

Since,

Area  = (width of interval) [sum of first and last ordinate /2 + sum of remaining ordinates]
                                  = 10\left[ {\frac{{0 + 0}}{2} + 7 + 11 + 15 + 5} \right] = 10 \times 38 = 380 square meters.

 

Example:

Apply the trapezoidal rule to find the area of a plot of land having the following dimensions:

Ordinates: 2,7,18,38 and 70m

Common distance: 33m

 

Solution:

Given that a = 2m, b = 7m, c = 18m, d = 38m, e = 70m respectively and S = 33m
\therefore by the trapezoidal rule:

Area  = S\left[ {\frac{{a + e}}{2} + \left( {b + c + d} \right)} \right]
 = 33\left[ {\frac{{2 + 70}}{2} + \left( {7 + 18 + 38} \right)} \right]
 = 33\left[ {36 + 63} \right] = 33\left( {99} \right) = 3102 square meters.