Calculate integral from 0 to 1 e^(-x²) dx using Simpson's

Calculate $\displaystyle\int_0^1e^{-x^2}\,dx$ using Simpson's method with accuracy $0.0001$.

Step 1: Fourth derivative

Differentiating $f(x)=e^{-x^2}$ four times gives $f^{(4)}(x)=(16x^4-48x^2+12)e^{-x^2}$. On $[0,1]$ this attains its largest magnitude at $x=0$: $f^{(4)}(0)=12$.

Step 2: Determine the number of subintervals

\[\frac{1}{180(2n)^4}\times12\le0.0001\ \Longrightarrow\ n^4\ge41.7\ \Longrightarrow\ n\ge3.\]

So $2n=6$ subintervals suffice.

Step 3: Apply Simpson's formula, $h=1/6$

Weighted sum $S\approx13.442947$, so

\[I_{\text{Simpson}}=\frac{1/6}{3}\times13.442947\approx0.746830.\]

\[\boxed{I_{\text{Simpson}}\approx0.746830}\]

Step 4: Accuracy check

A high-precision reference value is $I\approx0.746824$, giving actual error $\approx6.3\times10^{-6}$, comfortably inside the required $0.0001$ tolerance (and inside the theoretical bound, which works out to about $7.4\times10^{-5}$ for $2n=6$).

Step 5: Physical relevance

This Gaussian integral is directly related to the error function and the standard normal distribution: $\int_0^1e^{-x^2}dx=\tfrac{\sqrt\pi}{2}\,\mathrm{erf}(1)$. Because $e^{-x^2}$ is extremely smooth (infinitely differentiable, with rapidly decaying derivatives), Simpson's rule converges quickly, needing only 6 subintervals for 4-decimal accuracy.