Search any question & find its solution
Question:
Answered & Verified by Expert
By Simpson rule taking $n=4$, the value of the integral $\int_{0}^{1} \frac{1}{1+x^{2}} d x$ is equal to
Options:
Solution:
1299 Upvotes
Verified Answer
The correct answer is:
$0.785$
Here, $h=1 / 4,=0.25, y=\frac{1}{1+x^{2}}$
\begin{array}{c|c|c}
\hline & x & y \\
\hline 1 & 0 & 1.0 \\
2 & 0.25 & 0.941 \\
3 & 0.5 & 0.8 \\
4 & 0.75 & 0.64 \\
\hline 5 & 1 & 0.5 \\
\hline
\end{array}
$$
\begin{array}{c|c|c}
\hline & x & y \\
\hline 1 & 0 & 1.0 \\
2 & 0.25 & 0.941 \\
3 & 0.5 & 0.8 \\
4 & 0.75 & 0.64 \\
\hline 5 & 1 & 0.5 \\
\hline
\end{array}
$$
By Simpson's Rule
$$
\begin{array}{l}
\int_{0}^{1} \frac{d x}{1+x^{2}}=\frac{1}{4 \times 3} \\
\quad[(1+0.5)+4(0.941+0.64)+2(0.8)] \\
=\frac{1}{12}[9.424]=0.785
\end{array}
$$
\begin{array}{c|c|c}
\hline & x & y \\
\hline 1 & 0 & 1.0 \\
2 & 0.25 & 0.941 \\
3 & 0.5 & 0.8 \\
4 & 0.75 & 0.64 \\
\hline 5 & 1 & 0.5 \\
\hline
\end{array}
$$
\begin{array}{c|c|c}
\hline & x & y \\
\hline 1 & 0 & 1.0 \\
2 & 0.25 & 0.941 \\
3 & 0.5 & 0.8 \\
4 & 0.75 & 0.64 \\
\hline 5 & 1 & 0.5 \\
\hline
\end{array}
$$
By Simpson's Rule
$$
\begin{array}{l}
\int_{0}^{1} \frac{d x}{1+x^{2}}=\frac{1}{4 \times 3} \\
\quad[(1+0.5)+4(0.941+0.64)+2(0.8)] \\
=\frac{1}{12}[9.424]=0.785
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.