Question

By Simpson one third rule taking n=4, the value of the integral $\int\limits_0^1 {\dfrac{1}{{1 + {x^2}}}} dx$ ​ is equal to
A. 0.788
B. 0.781
C. 0.785
D. None of these

Answer 1

Hint: Simpson’s one third rule is used for integration of definite integrals when we divide a small interval [a, b] into two parts. After dividing the interval, we get; ${x_0}$=a, ${x_1}$= a + b, ${x_2}$ = b.

Complete Step-by-Step solution:
Hence, we can write the approximation as;
$\int\limits_a^b {f(x)dx} $ ≈ S2=$\dfrac{h}{3}\left[ {f({x_0}) + 4f({x_1}) + f({x_2})} \right]$
S2=$\dfrac{h}{3}\left[ {f(a) + 4f(\dfrac{{a + b}}{2}) + f(b)} \right]$
Where $h = \dfrac{{b - a}}{2}$
This is Simpson's one third rule for integration.
Given,
\[\begin{gathered}
  h = 41,0.25,y = \dfrac{1}{{1 + {x^2}}} \\
  \begin{array}{*{20}{l}}
  {}&{\mathbf{x}}&{\;{\text{ }}\;{\text{ }}\;{\mathbf{y}}} \\
  1&0&{1.0} \\
  2&{0.25\;\;{\text{ }}\;\;}&{0.941} \\
  {3\;}&{0.25}&{0.941} \\
  4&{0.75}&{0.64} \\
  5&1&{0.5}
\end{array} \\
\end{gathered} \]\[\]
By Simpson’s Rule
\[\int\limits_0^1 {\dfrac{{dx}}{{1 + {x^2}}}} = \dfrac{1}{{4 \times 3}}[(1 + 0.5) + 4(0.941 + 0.941 + 0.64) + 2(0.8)]\]
\[\begin{gathered}
   = \dfrac{1}{{12}}[9.424] \\
   = 0.785 \\
\end{gathered} \]
Hence, option (C) is the correct answer.

Note: Simpson's 1/3 rule is a method for numerical approximation of definite integrals. Specifically, it is the same as approximation In Simpson's 1/3 Rule, we use parabolas to approximate each part of the curve. we divide the area into (n) equal segments of width $\Delta x$.