Numerical Methods

Given $h = 1.0$, intervals: $x_0 = 0, x_1 = 1, x_2 = 2$. Function values $y = x^3$: $y_0 = 0^3 = 0$ $y_1 = 1^3 = 1$ $y_2 = 2^3 = 8$ Simpson's $1/3$ rule: $I = \frac{h}{3} [y_0 + 4y_1 + y_2] = \frac{1}{3} [0 + 4(1) + 8] = \frac{12}{3} = 4.0$ (Note: Simpson's rule integrates polynomials of degree up to 3 exactly).