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By the application of Simpson's one-third rule for numerical integration, with two subintervals, the value of $\int_{0}^{1} \frac{d x}{1+x}$ is
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Verified Answer
The correct answer is:
$\frac{25}{36}$
Since, the given integration is divided into two subintervals.
ie,
$$
h=\frac{1-0}{2}=\frac{1}{2}
$$
$$
\therefore \quad \int_{0}^{1} \frac{1}{1+x} d x=\frac{h}{3}\left[\left(y_{0}+y_{2}\right)+4\left(y_{1}\right)\right]
$$
At
$$
\begin{aligned}
x=0, y &=1 \\
x &=\frac{1}{2}, y_{1}=\frac{2}{3}
\end{aligned}
$$
and
$$
x=1, y_{2}=\frac{1}{2}
$$
$$
\begin{aligned}
\therefore \int_{0}^{1} \frac{1}{1+x} d x &=\frac{1}{2 \cdot 3}\left[\left(1+\frac{1}{2}\right)+4\left(\frac{2}{3}\right)\right] \\
&=\frac{1}{6}\left[\frac{3}{2}+\frac{8}{3}\right]=\frac{25}{36}
\end{aligned}
$$
ie,
$$
h=\frac{1-0}{2}=\frac{1}{2}
$$
$$
\therefore \quad \int_{0}^{1} \frac{1}{1+x} d x=\frac{h}{3}\left[\left(y_{0}+y_{2}\right)+4\left(y_{1}\right)\right]
$$
At
$$
\begin{aligned}
x=0, y &=1 \\
x &=\frac{1}{2}, y_{1}=\frac{2}{3}
\end{aligned}
$$
and
$$
x=1, y_{2}=\frac{1}{2}
$$
$$
\begin{aligned}
\therefore \int_{0}^{1} \frac{1}{1+x} d x &=\frac{1}{2 \cdot 3}\left[\left(1+\frac{1}{2}\right)+4\left(\frac{2}{3}\right)\right] \\
&=\frac{1}{6}\left[\frac{3}{2}+\frac{8}{3}\right]=\frac{25}{36}
\end{aligned}
$$
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