Calculate the approximate value of pi by applying

Calculate the approximate value of $\pi$ by applying Simpson's formula to $\displaystyle\int_0^1\frac{dx}{1+x^2}$, with $2n=10$, to an accuracy of $0.01$.

Step 1: Relate the integral to $\pi$

\[\int_0^1\frac{dx}{1+x^2}=\big[\arctan x\big]_0^1=\frac{\pi}{4}\quad\Longrightarrow\quad \pi=4\int_0^1\frac{dx}{1+x^2}.\]

Step 2: Simpson's formula, $2n=10$

$h=0.1$, nodes $x_i=0.1i$, weights $1,4,2,4,2,4,2,4,2,4,1$. Computing $f(x_i)=1/(1+x_i^2)$ at all 11 nodes and summing with weights gives weighted sum $S=23.561945$.

Step 3: Evaluate the integral

\[I_{\text{Simpson}}=\frac{0.1}{3}\times23.561945\approx0.785398.\]

Step 4: Recover $\pi$

\[\pi_{\text{approx}}=4\times0.785398\approx3.141599.\]

\[\boxed{\pi\approx3.1416}\]

Step 5: Accuracy check

\[E=|3.141593-3.141599|\approx4\times10^{-8}\ll0.01,\]

far exceeding the required accuracy of $0.01$ — in fact accurate to about 7 decimal places, since $f(x)=1/(1+x^2)$ is extremely smooth on $[0,1]$ and Simpson's rule converges rapidly for such functions.

Step 6: Theoretical error bound (for context)

With $\max_{[0,1]}|f^{(4)}|=24$ (attained at $x=0$), the bound on the integral's error is $\dfrac{1}{180\times10^4}\times24\approx1.33\times10^{-5}$, and on $\pi$ (multiplying by 4) about $5.3\times10^{-5}$ — itself already well under the required $0.01$, and the actual error turns out even smaller still.