Trapezoidal Rule with n=6 to approximate the integral of 1/(1+x^3) from x=0 to 3

Introduction : Trapezoidal rule will numerically approximate area under the curve 1/(1+x^3) from x=0 to 3. Let’s understand the rule step by step by solving example below

Question : Use Trapezoidal Rule with n=6 to approximate

\[\int\limits_0^3 {\frac{1}{{1 + {x^3}}}} dx\]

Solution :

Trapezoidal Rule with n =6 is given as

\[T_6 = {\textstyle{h \over 2}}\left( {f({x_0}) + 2f({x_1}) + 2f({x_2}) + 2f({x_3}) + 2f({x_4}) + 2f({x_5}) + f({x_6})} \right)\hspace{5 cm}(1)\]

where \(f(x_i) =\frac{1}{{1 + {x_i^3}}}, x_0=0, x_6=3\)

First we will calculate the size of the subinterval by dividing length of the interval [0, 3] by n =6.

\[h=\frac{3-0}{6}=\frac{3}{6}=0.5\]

In the table 1 values of grid points \(x_i=x_0+i*h\) for i =0,1,2,3,4,5,6 are presented

In the table 2 , function’s values at the grid points \(f(x_i )=\frac{1}{{1 + {x_i^3}}}\) for i =0,1,2,3,4,5,6 are presented

Now substituting required values in to the formula (1)

\[T_6 = \frac{0.5} { 2}\left( 1+ 2*0.888889 + 2*0.5+ 2*0.228571 + 2*0.111111 + 2*0.06015 +0.035714 \right) = 1.153289\]

Thus \(\int\limits_0^3 {\frac{1}{{1 + {x^3}}}} dx=1.153289\)