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MHT CET · Maths · Definite Integration

By Simpson rule taking \(n=4\), the value of the integral \(\int_{0}^{1} \frac{1}{1+x^{2}} d x\) is equal to

  1. A \(0.788\)
  2. B \(0.781\)
  3. C \(0.785\)
  4. D None of these
Verified Solution

Answer & Solution

Correct Answer

(C) \(0.785\)

Step-by-step Solution

Detailed explanation

Here, \(h=1 / 4,=0.25, y=\frac{1}{1+x^{2}}\)
\(x\)\(y\)
101.0
20.250.941
30.50.8
40.750.64
510.5
By Simpson's Rule
\(
\begin{array}{l}
\int_{0}^{1} \frac{d x}{1+x^{2}}=\frac{1}{4 \times 3} \\
\quad[(1+0.5)+4(0.941+0.64)+2(0.8)] \\
=\frac{1}{12}[9.424]=0.785
\end{array}
\)