The fundamental theorem of calculus is a central result in calculus connecting the derivative, which describes instantaneous change, with the integral, which describes accumulation. Its two complementary statements establish that integrating a continuous function produces an antiderivative, and that a definite integral can be evaluated by subtracting the endpoint values of an antiderivative. This connection replaces many calculations involving limits of sums with calculations involving functions and their derivatives. (openstax.org)
The two statements
Let (f:[a,b]\to\mathbb R) be a continuous function, where (a<b) are real numbers. Define the accumulation function [ A(x)=\int_a^x f(t),dt. ] The first statement says that (A) is continuous on ([a,b]), differentiable on ((a,b)), and satisfies [ A'(x)=f(x). ] Thus, (A) is an antiderivative of (f): a function whose derivative equals (f). The integration variable (t) is a dummy variable; (x) determines the moving upper endpoint. At the interval’s endpoints, the corresponding derivatives are understood one-sidedly. (openstax.org)
The second statement says that, if (F) is any antiderivative of (f), continuous on ([a,b]) and differentiable on ((a,b)), then [ \int_a^b f(x),dx=F(b)-F(a). ] The right-hand side is often written ([F(x)]_a^b). Adding a constant to (F) leaves this difference unchanged. Accordingly, integration reverses differentiation only up to an additive constant, while the condition (A(a)=0) selects a particular antiderivative. (math.mit.edu)
Why accumulation yields a derivative
The first statement has a direct proof using continuity. For an interior point (x) and sufficiently small nonzero (h), [ \frac{A(x+h)-A(x)}h =\frac1h\int_x^{x+h}f(t),dt. ] This expression is the average value of (f) over the small interval between (x) and (x+h). Subtracting (f(x)) gives the estimate [ \left|\frac{A(x+h)-A(x)}h-f(x)\right| \leq \sup_{t\text{ between }x\text{ and }x+h}|f(t)-f(x)|. ] Continuity makes the right-hand side tend to zero as (h\to0), so the defining limit of the derivative is (f(x)). Geometrically, a narrow strip contributes approximately its width multiplied by the function’s height; dividing by the width recovers that height in the limit. (math.mit.edu)
The second statement follows because (F-A) has derivative zero. By the mean value theorem, it is constant on the interval. Consequently, (F(b)-F(a)=A(b)-A(a)), which equals the required integral. (ocw.mit.edu)
Evaluation and variable limits
For the polynomial (f(x)=x^2), an antiderivative is (F(x)=x^3/3). Hence [ \int_0^2x^2,dx =\left[\frac{x^3}{3}\right]_0^2 =\frac83. ] This illustrates the practical distinction between defining an integral through Riemann sums and evaluating it through an antiderivative. The theorem does not redefine integration as antidifferentiation; it proves that these independently introduced operations agree under suitable hypotheses. (math.mit.edu)
Combining the theorem with the chain rule handles moving endpoints. If (u) and (v) are differentiable and (f) is continuous on an interval containing their values, then [ \frac{d}{dx}\int_{u(x)}^{v(x)}f(t),dt =f(v(x))v'(x)-f(u(x))u'(x). ] For example, [ \frac{d}{dx}\int_0^{x^2}\cos t,dt =2x\cos(x^2). ] Only the endpoints depend on (x) in this formula. Integrands that themselves depend on an additional parameter require separate results on differentiation under the integral sign. (openstax.org)
Net change and applications
The endpoint formula also expresses a net-change principle: [ \int_a^b Q'(t),dt=Q(b)-Q(a). ] When (Q') is continuous, integrating a rate of change gives the total change in the underlying quantity. In classical mechanics, integrating velocity gives displacement, whereas integrating speed gives distance traveled. These differ when the direction of motion changes. Similarly, positive and negative contributions to an integral can cancel; an integral representing signed area need not equal the total geometric area between a curve and an axis. (openstax.org)
The theorem also connects differential and integral formulations of differential equations. With continuous (f), the conditions (y'=f) and (y(a)=c) are satisfied by [ y(x)=c+\int_a^x f(t),dt. ] This formula separates the accumulated change from the initial value. (math.mit.edu)
Hypotheses and extensions
Continuity is a convenient sufficient hypothesis, not the weakest possible one. If (f) is Riemann integrable, its accumulation function is differentiable with derivative (f(x)) at every interior point where (f) is continuous. The endpoint formula also holds for a Riemann-integrable function possessing an antiderivative with the stated endpoint continuity. (ocw.mit.edu)
In mathematical analysis, the Lebesgue integral supports a broader formulation: a function (F) on a compact interval is absolutely continuous precisely when it can be represented as an initial value plus the integral of an integrable function. In that case, (F') exists almost everywhere, is integrable, and [ F(x)-F(a)=\int_a^x F'(t),dt. ] Here “almost everywhere” allows an exceptional set of Lebesgue measure zero. (arxiv.org)
In higher dimensions, Stokes’ theorem extends the same relationship between interior change and boundary values: integration of an appropriate derivative over a region becomes integration over its boundary. The one-dimensional endpoint difference is the simplest instance of this principle. (openstax.org)