Examples Using Trapezoidal Rule

Example 1: 

1. Find the area under the curve using trapezoidal rule formula which passes through the following points:x00.511.5y56911

Solution:

Given:

y₀ = 5
y₁= 6
y₂= 9
y₃= 11
h = (0.5 – 0) = (1 – 0.5) = (1.5 – 1) = 0.5

Using Trapezoidal rule formula,Area = (h/2) [y₀ + y_n + 2(y₁ + y₂ + y₃ + ….. + y_n-1)]

=(.5/2) [5 + 11 + 2 (6 + 9)]

= 0.25 [16+30]= 0.25 [46]= 11.5

Answer: 

Therefore, the area under the curve is 11.5 sq units.

Example 2:  

Using Trapezoidal Rule Formula find the area under the curve y = x² between x = 0 and x = 4 using the step size of 1.

Solution:

Given:

y = x²
h = 1

Find the values of ‘y’ for different values of ‘x’ by putting the value of ‘x’ in the equation

y = x²X01234y = x₂y₀ = 0y₁ = 1y₂ = 4y₃ = 9y₄ = 16

Using Trapezoidal rule:

Area = (h/2) [y₀ + y_n + 2 (y₁ + y₂ + y₃ + ….. + y_n-1)]

= (1/2) [0 + 16 + 2 (1 + 4 + 9)]= 0.5 [16 + 28]

= 22

Answer: 

Therefore, the area under the curve is 22 sq units.

Example 3: 

Find the area under the curve using the trapezoidal rule formula which passes through the following points:

x00.511.5

y471015

 Solution: 

Given: y₀ = 4
y₁ = 7
y₂ = 10
y₃ = 15
h = (0.5 – 0) = (1 – 0.5) = (1.5 – 1) = 0.5

Using Trapezoidal formula:

Area = (h/2) [y₀ + y_n + 2 (y₁ + y₂ + y₃ + ….. + y_n-1)]= (0.5/2) [4 + 15 + 2 (7 + 10)]

= 0.25 [19 + 34]= 0.25 [53]

= 13.25

Answer: 

Therefore, the area under the curve is 13.25 sq units.