Calculate integral from 1 to ²(x³-x) dx using Simpson's

Calculate $\displaystyle\int_1^2(x^3-x)\,dx$ using Simpson's formula and estimate the error.

Step 1: Exact value

\[I_{\text{exact}}=\int_1^2(x^3-x)\,dx=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_1^2=\left(4-2\right)-\left(\frac14-\frac12\right)=2+\frac14=\frac94=2.25.\]

\[\boxed{I_{\text{exact}}=2.25}\]

Step 2: Simpson's formula with $n=2$

$h=0.5$, nodes $x_0=1,\ x_1=1.5,\ x_2=2$: $f(1)=0$, $f(1.5)=1.875$, $f(2)=6$.

\[I_{\text{Simpson}}=\frac{0.5}{3}\left[0+4(1.875)+6\right]=\frac{0.5}{3}\times13.5=2.25.\]

\[\boxed{I_{\text{Simpson}}=2.25}\]

Step 3: Error

\[\text{Error}=|I_{\text{exact}}-I_{\text{Simpson}}|=0.\]

Step 4: Why the error is exactly zero

Simpson's rule is exact for cubic polynomials (Problem 198). Here $f(x)=x^3-x$ has fourth derivative $f^{(4)}(x)\equiv0$, and the Simpson error formula $E=-\dfrac{(b-a)^5}{180n^4}f^{(4)}(\xi)$ vanishes identically regardless of $\xi$ or $n$, guaranteeing zero error for any valid $n\ge2$.

Step 5: Verification with $n=4$

Repeating with $h=0.25$ and nodes $1,1.25,1.5,1.75,2$ gives $\tfrac{0.25}{3}[0+4(0.703125)+2(1.875)+4(3.609375)+6]=\tfrac{0.25}{3}\times27=2.25$ again — confirming the result is independent of $n$, exactly as the vanishing fourth derivative predicts.