FTC Part 2 (Evaluation)

Real Analysis · Axiom Academy

The Fundamental Theorem of Calculus, Part 2 — turn a hard limit of Riemann sums into a single subtraction of antiderivative values. If f is continuous on [a,b] and F is any antiderivative of f — meaning F'(x) = f(x) — then the definite integral is computed by evaluating F at the two endpoints and subtracting. the signed area under f from a to b … … equals the net change of an antiderivative F So there are no Riemann sums to take — just three small jobs: 2. Why It Works: Area Is the Net Change of F Part 1 of the theorem says the accumulated-area function is itself an antiderivative of f . Any other antiderivative F differs from A only by a constant , so they have the same net change across [a,b] — and A 's net change is exactly the area we want. The antiderivative's value at the upper limit: the height of F at x = b . Its value at the lower limit: the height of F at x = a . The vertical change in F from a to b — and that change equals the shaded area under f . If f is a rate of change, integrating it recovers the total change in the quantity it drives. Net change: velocity to displacement If v(t) is velocity and s(t) is position, then s'(t) = v(t) , so v is the rate of change of s . The Evaluation Theorem then reads — the integral of velocity is the net displacement . For example, with v(t) = 2t on [1,4] an antiderivative is s(t) = t^2 , giving a displacement of s(4) - s(1) = 16 - 1 = 15 .

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