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Let $f(0)=1, f(0.5)=\frac{5}{4}, f(1)=2, f(1.5)=\frac{13}{4}$ and $f(2)=5$. Using Simpson's rule, $\int_0^2 f(x) d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{14}{3}$
$\begin{aligned}
& \because h=\frac{b-a}{n} \\
& \Rightarrow \quad h=\frac{2-0}{4}=0.5
\end{aligned}$
$\therefore$ Simpson's rule,
$\begin{aligned} \int_a^b f(x) d x= & \frac{h}{3}\left[\left(y_0+y_4\right)+4\left(y_1+y_3\right)+2\left(y_2\right)\right] \\ & =\frac{0.5}{3}\left[(1+5)+4\left(\frac{5}{4}+\frac{13}{4}\right)+2(2)\right] \\ & =\frac{0.5}{3}[6+18+4]=\frac{14}{3}\end{aligned}$
& \because h=\frac{b-a}{n} \\
& \Rightarrow \quad h=\frac{2-0}{4}=0.5
\end{aligned}$
$\therefore$ Simpson's rule,
$\begin{aligned} \int_a^b f(x) d x= & \frac{h}{3}\left[\left(y_0+y_4\right)+4\left(y_1+y_3\right)+2\left(y_2\right)\right] \\ & =\frac{0.5}{3}\left[(1+5)+4\left(\frac{5}{4}+\frac{13}{4}\right)+2(2)\right] \\ & =\frac{0.5}{3}[6+18+4]=\frac{14}{3}\end{aligned}$
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