Fundamental Theorem of Calculus

The central idea

If $f$ is continuous on $[a,b]$ and $F(x)=\int_a^x f(t)dt$, then $F'(x)=f(x)$ for $a<x<b$ (FTC Part I). If $G$ is any antiderivative of $f$, then $\int_a^b f(x)dx=G(b)-G(a)$ (FTC Part II). The dummy variable $t$ prevents confusion between the integration variable and moving endpoint.

Definitions, hypotheses, and notation

Part I requires enough regularity for the accumulation function to inherit the integrand as its derivative at the point. Continuity is the standard introductory hypothesis. Part II requires an antiderivative across the interval and turns a global accumulation into endpoint data. Constants vanish because $(G(b)+C)-(G(a)+C)=G(b)-G(a)$.

With variable endpoints, orientation explains the signs. Increasing an upper limit adds a thin strip, whereas increasing a lower limit removes one. Thus $d/dx\int_{u(x)}^{v(x)}f(t)dt=f(v(x))v'(x)-f(u(x))u'(x)$. This formula combines FTC Part I, the chain rule, and the convention that reversing integration bounds negates the integral.

Conceptual meaning

Accumulating a continuous rate and then measuring how fast the accumulation changes returns the current rate. Conversely, endpoint subtraction converts infinitely many small contributions into an exact total. The theorem makes differentiation and integration inverse processes, with constants canceled at endpoints.

A dependable method and decision rule

1. Decide whether the task differentiates an accumulation function or evaluates an integral.
2. For a moving upper limit $g(x)$, use $d/dx\int_a^{g(x)}f(t)dt=f(g(x))g'(x)$.
3. For two moving limits, subtract the lower-endpoint contribution.
4. For a definite integral, find an antiderivative and compute upper minus lower.
5. Check continuity or the appropriate weaker integrability assumptions before invoking the theorem.

Fully worked example

Common mistakes and why they fail