FTC Part 2 - Evaluation of Integrals

Calculus 1 · Axiom Academy

LESSON FTC Part 2: Evaluation of Integrals To find the area under a curve, just evaluate an antiderivative at the two endpoints and subtract. 1. The Fundamental Theorem (Part 2) If f is continuous on [a, b] and F is any antiderivative of f (that is, F' = f ), then the definite integral is just a difference of two numbers: Find an antiderivative, plug in the endpoints, subtract. Here is the idea behind the formula. If F' = f , then f is the rate of change of F . The integral adds up all the tiny changes from a to b — and the sum of all those little changes is just the total change in F . Each thin strip of area under the rate f is one small step the antiderivative takes; stacking them from a to b carries F from F(a) up to F(b) . Let's evaluate . The integrand is f(x) = 2x , so we need an antiderivative — a function whose derivative is 2x . The blue region is a trapezoid of area 15 — exactly the gap between the heights of F(x) = x^2 at x = 4 and x = 1 . Area below, height-difference above: the same number. To keep the bookkeeping clear, mathematicians write the antiderivative in a bracket with the limits on the right. The bar means "plug in the top, then subtract the bottom": The positive area before and the negative area after it cancel exactly, so the net signed area is 0 . Evaluate the antiderivative at b — this is the F(b) term. Subtract its value at a — the F(a) term.

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