Approximating Definite Integrals — Question 7

Question 7

Show that Simpson’s Rule with $n=2$ is exact for $$\int_0^2x^3\,dx.$$

Question 7 – Solution

Step 1: Find the width and nodes. $$h=\frac{2-0}{2}=1,\qquad x_0=0,\ x_1=1,\ x_2=2.$$ Step 2: Apply Simpson’s Rule. $$\begin{align*} S_2&=\frac h3[f(x_0)+4f(x_1)+f(x_2)]\\ &=\frac13[0^3+4(1^3)+2^3]\\ &=\frac13(0+4+8)=4. \end{align*}$$ Step 3: Evaluate the exact integral. $$\begin{align*} I&=\int_0^2x^3\,dx =\left[\frac{x^4}{4}\right]_0^2\\ &=\frac{16}{4}-0=4. \end{align*}$$ Since $S_2=I$, Simpson’s Rule is exact here. $$\boxed{S_2=I=4}$$