FTC Part 2 (the Evaluation Theorem) states that if F is any antiderivative of f on [a, b], then the integral from a to b of f(x) dx = F(b) - F(a). This transforms the problem of computing a definite integral from a limit of Riemann sums (hard) into finding an antiderivative and evaluating at the endpoints (often easy). This is the most computationally powerful theorem in introductory calculus.
Evaluate definite integrals using the notation F(x) evaluated from a to b = F(b) - F(a). Practice with polynomial, trigonometric, and exponential integrands. Compare with Riemann sum approximations to verify. Emphasize that the +C cancels out in definite integrals.
From FTC Part 1, you learned that the accumulation function A(x) = ∫_a^x f(t) dt is an antiderivative of f — differentiating the integral recovers the integrand. FTC Part 2 is the payoff: it gives you a practical method to *evaluate* that integral. If F is any antiderivative of f on [a, b] — meaning F'(x) = f(x) — then ∫_a^b f(x) dx = F(b) − F(a). The definite integral, which was defined as a limit of Riemann sums (a painful computation), reduces to two function evaluations and a subtraction.
To see why this works, recall from Part 1 that A(x) = ∫_a^x f(t) dt is one antiderivative of f. Any other antiderivative F of f differs from A by a constant: F(x) = A(x) + C for some C. Now compute F(b) − F(a): you get [A(b) + C] − [A(a) + C] = A(b) − A(a). Since A(a) = ∫_a^a f(t) dt = 0, this simplifies to A(b) = ∫_a^b f(t) dt. The constant C cancels regardless of which antiderivative you choose — this is why you do not write "+C" when evaluating definite integrals, and why any correct antiderivative gives the same answer.
In practice, the standard notation is [F(x)]_a^b = F(b) − F(a), evaluated after finding F. For example, ∫_0^3 x² dx = [x³/3]_0^3 = 27/3 − 0 = 9. No limits, no rectangles — just one antiderivative and two evaluations. This efficiency is what makes the theorem so powerful. Every rule you know for finding antiderivatives (power rule, trig integrals, exponentials) becomes a tool for evaluating definite integrals. The upcoming techniques of u-substitution and integration by parts will further extend your antiderivative toolkit, and FTC Part 2 converts each new technique directly into a method for computing areas, accumulated quantities, and net change.
One subtlety: the theorem requires f to be continuous (or nearly so) on [a, b]. If f has a jump discontinuity inside the interval, the antiderivative F may not be differentiable at that point, and the simple subtraction formula can give a wrong answer. When you encounter piecewise or discontinuous integrands, split the integral at the discontinuity and apply FTC Part 2 to each piece separately.