The second half turns the theorem into a method. If any antiderivative of the integrand can be found, the integral is the difference of its values at the two ends, and no partition is ever formed. The proof is the Mean Value Theorem applied on each piece of a partition, and it needs no continuity of the integrand at all — only that the integrand is integrable and is the derivative of something.

A journey’s total distance can be found by adding up the distances covered in each minute, or by reading the odometer at the start and at the end. The theorem says the two agree, and the second is the one anybody actually uses.

Suppose $f$ is integrable on $[a,b]$ and $G$ is a function with $G'=f$ throughout $[a,b]$. Then $\int_a^b f=G(b)-G(a)$. Note what is not assumed: $f$ need not be continuous. Note also what follows: any two antiderivatives differ by a constant, so the choice of $G$ does not matter.

• Any antiderivative serves: two of them differ by a constant on an interval, and the constant cancels in the difference.
• The theorem does not say an antiderivative exists; the first half does, for a continuous integrand.
• Substitution and integration by parts are the chain rule and the product rule read through this theorem.
• Most integrands have no antiderivative in terms of familiar functions, and then this theorem is of no help; the integral still exists.
• The two halves are different statements: the first differentiates an integral, the second integrates a derivative.

Take any partition and telescope. On each piece apply the Mean Value Theorem to the antiderivative: its change across the piece is the derivative at some interior point times the width, and that derivative is the integrand. So the difference of the end values is a Riemann sum of the integrand, with tags supplied by the theorem rather than chosen. That sum is trapped between the lower and upper sums of the partition, so the difference of the end values lies in every such bracket — and the integral is the only number that does.

Take an arbitrary partition of the interval.

The sum telescopes: every interior value is added once and subtracted once.

The Mean Value Theorem on each piece, with the derivative of G being f.

So the difference of the end values is a Riemann sum with these tags.

Every Riemann sum is trapped by its partition.

One number lies in every such bracket, and integrability says it is the integral.

Any function whose derivative is the integrand serves, and the constant cancels.

• An antiderivative evaluates an integral as a difference of two values.
• Any antiderivative serves; the constant cancels.
• The integrand need not be continuous.
• The proof telescopes and applies the Mean Value Theorem on each piece.

Great work! You now know how to:

• Evaluate an integral from an antiderivative
• Say what the theorem assumes and what it does not
• Keep the two halves of the Fundamental Theorem apart
• Say why the choice of antiderivative does not matter