Fundamental Theorem of Calculus

Analysis, Measure & PDE · Undergraduate · Mathematics Resources editorial compilation (AI-assisted)

For continuous integrands, differentiation recovers accumulation and antiderivatives evaluate definite integrals.

Overview

The fundamental theorem of calculus connects a local operation, differentiation, with a global operation, integration. Its two directions answer different questions: what is the rate of change of an accumulated quantity, and how can total accumulation be computed from a primitive? Keeping these directions distinct prevents a computational formula from being applied without its hypotheses.

The theorem explains why an integral can often be evaluated by finding a function whose derivative is the integrand. It does not assert that every useful integrand has an elementary primitive.

Prerequisites

Understand derivatives, continuous functions, and the Riemann integral on a compact interval. An antiderivative of $f$ is a differentiable function $F$ satisfying $F'=f$ . Two antiderivatives on an interval differ by a constant, a consequence of the mean value theorem.

Definition and conventions

Let $a<b$ and $f:[a,b]\to\mathbb R$ be continuous. Define

A(x)=\int_a^x f(t)\,dt.

Then $A$ is differentiable on $(a,b)$ and $A'(x)=f(x)$ . Moreover, if $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $F'=f$ , then

\int_a^b f(t)\,dt=F(b)-F(a).

Endpoint derivatives, when needed, are one-sided. Reversing the integration bounds changes the sign. These are classical continuous-integrand statements; more general versions need different assumptions.

Intuition

Adding a short interval of width $h$ changes accumulation by approximately $hf(x)$ . Dividing by $h$ gives an average of nearby function values. Continuity forces that average toward $f(x)$ , which is why accumulation differentiates back to the integrand.

Examples and non-examples

For $f(t)=3t^2$ on $[0,2]$ , the accumulation is $A(x)=x^3$ , so $A'(x)=3x^2$ and the total integral is $8$ .

Let $f$ be zero for negative inputs and one for nonnegative inputs on $[-1,1]$ . Its accumulation from $-1$ is zero up to zero and equals $x$ afterward. The left and right derivatives at zero are $0$ and $1$ . Thus the formula $A'=f$ fails at the jump, although $f$ is Riemann integrable. Integrability alone does not give pointwise differentiability everywhere.

Important properties

For nonzero $h$ near an interior point,

\frac{A(x+h)-A(x)}h-f(x)=\frac1h\int_x^{x+h}(f(t)-f(x))\,dt.

Its absolute value is bounded by the largest nearby deviation of $f(t)$ from $f(x)$ , which tends to zero. This proves the first direction. Subtracting $A$ from $F$ gives a function with zero derivative; the mean value theorem makes it constant, proving the second.

Integration by parts follows by applying the theorem to a product of continuously differentiable functions. Substitution follows from the chain rule with the appropriate interval and orientation conditions.

Connections and further reading

Lebesgue integration replaces continuity by integrability in an almost-everywhere differentiation theorem. Recovering a function from its derivative requires absolute continuity, rather than merely an almost-everywhere derivative. Taylor, §4.2 proves the classical statements used here.