Question

Using Simpson’s one-third rule (with step size $h=0.25$), the area under the curve $y = e^{-x^{3}}$, from $x=0$ to $x=1$ is ............. (rounded to two decimal places).

Hint:

For equally spaced nodes $x_0,\dots,x_4$ with $n=4$ (even), Simpson’s $1/3$ uses weights $1,4,2,4,1$. Compute odd-index sum, even-index sum, then plug into $\frac{h}{3}[\,\cdot\,]$.

Answer 1

Step 1: Tabulate values with $h=0.25$.
$x: \; 0,\;0.25,\;0.5,\;0.75,\;1$; \quad $y=e^{-x^3}:\; f_0=1,\; f_1=e^{-0.25^3},\; f_2=e^{-0.5^3},\; f_3=e^{-0.75^3},\; f_4=e^{-1}$.
Numerically: $f_0=1.0000,\; f_1=0.9845,\; f_2=0.8825,\; f_3=0.6558,\; f_4=0.3679$.
Step 2: Apply Simpson’s $1/3$ rule.
$\displaystyle I \approx \frac{h}{3}\Big[f_0+f_4+4(f_1+f_3)+2(f_2)\Big]$
$\displaystyle \Rightarrow I \approx \frac{0.25}{3}\Big[1+0.3679+4(0.9845+0.6558)+2(0.8825)\Big]$
$\displaystyle \Rightarrow I \approx 0.8078 \approx 0.81\ (\text{to two decimals}).$