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

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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

x_i00.511.522.53
Table 1

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

f(x_i)10.8888890.50.2285710.1111110.060150.035714
Table 2

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


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