Fundamental Theorem of Calculus Part II: How to Evaluate Integrals

The fundamental theorem of calculus part ii gives you the practical shortcut for evaluating a definite integral: find any antiderivative of the integrand, then subtract its value at the lower limit from its value at the upper limit. In symbols, if F is an antiderivative of f, then the integral of f from a to b equals F(b) − F(a). This turns what could be an infinite limiting process into two substitutions and one subtraction — which is why it is the workhorse of every calculus course.

The Statement of the Theorem

Let f be continuous on the closed interval [a, b], and let F be any function whose derivative is f on that interval. Then:

∫ from a to b of f(x) dx = F(b) − F(a)

A few details matter here. First, F is called an antiderivative of f, and the theorem requires only that F′ = f. Second, the choice of F does not change the answer. If you switch to F + C for some constant C, the terms F(b) + C and F(a) + C cancel in the subtraction, leaving F(b) − F(a) exactly as before. That is why we drop the “+ C” when evaluating definite integrals.

The result is often written using the evaluation bar notation: [F(x)] from a to b, shorthand for F(b) − F(a). The theorem is the natural counterpart to the first part, which says that integrating a continuous function up to a variable upper bound produces an antiderivative. Together they show that differentiation and integration are inverse operations in a precise sense (Wikipedia).

A Worked Example

Evaluate ∫ from 1 to 3 of x² dx.

The integrand is f(x) = x². An antiderivative is F(x) = x³/3, since the derivative of x³/3 is x².

Now apply the limits:

F(3) − F(1) = 3³/3 − 1³/3 = 27/3 − 1/3 = 26/3

So the integral equals 26/3, roughly 8.67 — the exact area under the parabola y = x² between x = 1 and x = 3. Notice how little calculation was needed: no rectangles, no limits, no summation formulas. The theorem converted a geometric area problem into arithmetic.

Contrast this with the numerical approach, where you would slice the interval into thin strips, add their areas, and take a limit as the strips shrink. That method still works and is the definition of the integral, but it is far more laborious for a simple polynomial. The theorem exists precisely to let you skip it (Khan Academy).

When the Theorem Applies — and Its Limits

The continuity condition is not decorative. The theorem guarantees the result when f is continuous on [a, b]. If f has a discontinuity inside the interval, the integral may fail to exist, or an antiderivative may not behave as expected across the jump. A classic trap is integrating 1/x² from −1 to 1; a naive antiderivative application gives a nonsensical answer because the function blows up at x = 0, which lies inside the interval.

Even with these caveats, the theorem is the standard route for any definite integral you can antidifferentiate symbolically — and when no elementary antiderivative exists, numerical methods take over (Paul’s Online Math Notes, MIT OpenCourseWare).

A useful habit: after computing a definite integral, sanity-check the sign. If f is positive throughout [a, b], the integral must be positive. If f crosses the axis, the integral measures net signed area, so cancellation is expected and a small answer is not automatically wrong.

Key Takeaways

• The second part of the fundamental theorem states that ∫ from a to b of f(x) dx = F(b) − F(a), where F is any antiderivative of f.
• The constant of integration never matters for a definite integral, because it cancels in the subtraction.
• Evaluating a definite integral is a three-step routine: find an antiderivative, substitute the two limits, subtract.
• Reversing the limits flips the sign, and continuity of f on [a, b] is required for the theorem to apply cleanly.
• The theorem reduces area problems to arithmetic; when no elementary antiderivative exists, numerical integration is the fallback.