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Question: Considering six sub-interval, the value of \(\int_{0}^{6}\frac{dx}{1 + x^{2}}\) by Simpson's rule is...

Considering six sub-interval, the value of ∫06dx1+x2\int_{0}^{6}\frac{dx}{1 + x^{2}} by Simpson's rule is

A

1.3562

B

1.3662

C

1.3456

D

1.2662

Answer

1.3662

Explanation

Solution

h=6−06=1h = \frac{6 - 0}{6} = 1

x0=0, 6mu 6mux1=0+1.1=1x_{0} = 0,\mspace{6mu}\mspace{6mu} x_{1} = 0 + 1.1 = 1

x2=0+2.1=2, 6mu 6mux3=3, 6mu 6mux4=4, 6mu 6mux5=5, 6mu 6mux6=6x_{2} = 0 + 2.1 = 2,\mspace{6mu}\mspace{6mu} x_{3} = 3,\mspace{6mu}\mspace{6mu} x_{4} = 4,\mspace{6mu}\mspace{6mu} x_{5} = 5,\mspace{6mu}\mspace{6mu} x_{6} = 6and

}{\frac{1}{26},\mspace{6mu}\mspace{6mu} y_{6} = \frac{1}{37}}$$ By Simpson's rule, $$\int_{a}^{b}{f(x)dx = \frac{h}{3}\left\lbrack (y_{0} + y_{6}) + 4(y_{1} + y_{3} + y_{5}) + 2(y_{2} + y_{4}) \right\rbrack} = \frac{1}{3}\left\lbrack \left( 1 + \frac{1}{37} \right) + 4\left( \frac{1}{2} + \frac{1}{10} + \frac{1}{26} \right) + 2\left( \frac{1}{5} + \frac{1}{17} \right) \right\rbrack = 1.3662$$