Question

By Simpson's rule taking $n = 4$, the value of the integral $\int_{0}^{1}{\frac{1}{1 + x^{2}}dx}$ is equal to

• A. 0.785
• B. 0.788
• C. 0.781
• D. None of these

Answer 1

Answer: (A)

0.785

Answer 2

$h = \frac{1 - 0}{4} = 0.25$

$x_{0} = 0\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu} x_{1} = 0 + 0.25 = 0.25$

$x_{2} = 0 + 2(0.25) = 0.50,\mspace{6mu}\mspace{6mu}\mspace{6mu} x_{3} = 0.75,\mspace{6mu}\mspace{6mu} x_{4} = 1$and

$y_{0} = \frac{1}{1 + x_{0}^{2}} = \frac{1}{1 + 0} = 1$,$y_{1} = \frac{1}{1 + (0.25)^{2}} = 0.941$,

$y_{2} = \frac{1}{1 + (0.50)^{2}} = 0.8$, $y_{3} = 0.64$ and $y_{4} = 0.5$

$\int_{0}^{1}{\frac{dx}{1 + x^{2}} = \frac{0.25}{3}\left\lbrack (y_{0} + y_{4}) + 4(y_{1} + y_{3}) + 2(y_{2}) \right\rbrack}$

$= \frac{0.25}{3}\left\lbrack (1 + 0.5) + 4(0.941 + 0.64) + 2(0.8) \right\rbrack = 0.785$