Consider the below data:

\(\begin{array}{*{20}{c}} x&:&0&1&2\\ {f\left( x \right)}&:&4&3&{12} \end{array}\)

The value of\(\mathop \smallint \nolimits_0^2 f\left( x \right)dx\) by Trapezoidal rule will be:

Concept:

Trapezoidal rule states that for a function y = f(x)

xn = x0 + nh, where n = Number of sub-intervals

h = step-size

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + {y_3} + \ldots + {y_{n - 1}}} \right)} \right]\)   ---(1)

For a trapezoidal rule, a number of sub-intervals must be a multiple of 1.

Calculation:

\(\begin{array}{*{20}{c}} x&:&0&1&2\\ {f\left( x \right)}&:&4&3&{12} \end{array}\)

Here: x₀ = 4, x₁ = 3, x₂ = 12, h = 1

From equation (1);

\(\mathop \smallint \limits_0^2 f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{x_0} + {x_2}} \right) + 2\left( {{x_1}} \right)} \right]\)

\( = \frac{1}{2}\left[ {\left( {{4} + {12}} \right) + 2\left( {{3}} \right)} \right]={22\over2}=11\)

Key Points:

Apart from the trapezoidal rule, other numerical integration methods are:

Simpson’s one-third rule:

For applying this rule, the number of subintervals must be a multiple of 2.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots + {y_{n - 2}}} \right)} \right]\)     ..2)

Simpson’s three-eighths rule:

For applying this rule, the number of subintervals must be a multiple of 3.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{{3h}}{8}\left[ {\left( {{y_0} + {y_n}} \right) + 3\left( {{y_1} + {y_2} + {y_4} + {y_5} + \ldots } \right) + 2\left( {{y_3} + {y_6} + \ldots } \right)} \right]\)