Fundamental Theorem of Calculus Part 2

Calculus 1 · Axiom Academy

LESSON The Fundamental Theorem, Part 2 The bridge between antiderivatives and area: evaluate a definite integral by plugging the endpoints into any antiderivative. 1. The Theorem: Area Becomes Evaluation Suppose f is continuous on [a, b] and F is any antiderivative of f — meaning F' = f . Then the area captured by the integral equals a single subtraction. Find any antiderivative, then subtract its endpoint values. Why should area equal a subtraction? Because f is the rate of change of F . Adding up every infinitesimal change of F from a to b — which is exactly what does — must total the net change in F across the interval. Heights on the antiderivative F(x) = x^2 : it starts at F(1) = 1 and ends at F(4) = 16 . Since F' = f , the integral sums all the tiny rises that build F from a to b . Final minus initial height: F(4) - F(1) = 16 - 1 = 15 — the same 15 as the area. If f is velocity, F is position: is the net displacement, final position minus initial. The shaded area under f in Step 1 and the vertical net-change bracket on F here are both 15 . That equality, made precise, is the whole theorem. In practice we use a compact bracket notation. The vertical bar with limits means: evaluate the antiderivative at the top, then subtract its value at the bottom. Find an antiderivative of f(x) = 2x . Since , take F(x) = x^2 . Write it with the bracket and plug in the limits: . Subtract: F(4) - F(1) = 16 - 1 = 15 .

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