Integrals: Areas, Averages & Numerical Rules

The exam rarely asks for a clever antiderivative. It asks for an area, an average over an interval, or an estimate from a handful of tabulated values — the way a cross-section or a hydrograph actually arrives.

Worked: one curve, three ways

f(x) = 3x² − 2x + 1 on [0, 2]. Exactly: F(x) = x³ − x² + x, so the area is 8 − 4 + 2 = 6, and the average value is 6/2 = 3.

From the three values f(0) = 1, f(1) = 2, f(2) = 9 with h = 1: the trapezoidal rule gives ½(1 + 4 + 9) = 7, overestimating a curve that bends upward. Simpson gives ⅓(1 + 8 + 9) = 6 — exact, as promised for a quadratic.