Fundamental theorem of calculus

The theorem comes in two parts. The second part (often taught first) says that if F is any antiderivative of a continuous function f, then the definite integral of f from a to b equals F(b) − F(a). This is what turns integration from an infinite limiting process into an arithmetic step: instead of summing rectangles forever, you find an antiderivative and subtract.

The first part says that if you define an accumulation function g(x) = ∫ from a to x of f(t) dt, then g′(x) = f(x). In words, differentiating an integral with a variable upper limit returns the original integrand. When the upper limit is itself a function, the chain rule applies: the derivative of ∫ from a to u(x) of f(t) dt is f(u(x)) · u′(x).

A concrete evaluation shows part two at work. To compute ∫ from 1 to 3 of 2x dx, take the antiderivative x², then evaluate: 3² − 1² = 9 − 1 = 8. Part one is what justifies the whole procedure — it establishes that the accumulated-area function really is an antiderivative, connecting the geometric idea of area under a curve to the algebraic idea of reversing a derivative.

The theorem also underlies the interpretation of integrals in applied problems: if a function gives a rate of change, its definite integral gives the net change over an interval. That is why integrating velocity yields displacement while integrating speed yields total distance traveled. AP Calculus AB tests both parts heavily — evaluating definite integrals, differentiating accumulation functions with the chain rule, and applying net change to motion problems on free-response questions.