Definite Integral

Cashing in the Fundamental Theorem: find an antiderivative, subtract.

The idea

You already know what $\int_{a}^{b} f(x) \, dx$ means, for two numbers $a$ and $b$: the number those slice sums close in on. And you know the Fundamental Theorem promises you never have to work that limit out by hand.

Theorem (Fundamental Theorem of Calculus).

If $f$ is continuous on $[a, b]$ and $F$ is any function whose derivative is $f$, then $\int_{a}^{b} f(x) \, dx = F(b) - F(a).$

So the whole job comes down to one question: which function has $f$ as its derivative? Such an $F$ is called an antiderivative of $f$. Find one, evaluate it at both ends, subtract.

For powers the reversal is easy to guess. Differentiating $x^{n+1}$ gives $(n+1)x^{n}$, so to undo that you raise the exponent by one and divide by the new exponent.

Proposition (Power rule for antiderivatives).

For every $n \neq -1$, $\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C.$

An antiderivative is never unique — adding a constant changes no derivative — but here the constant does not matter: it appears in both $F(b)$ and $F(a)$ and cancels in the subtraction.

Ways to work on it

• Walkthrough. Antiderivatives, and using them to evaluate an integral.
• Practice. Random _a^b x^n dx — antiderivative, plug in, subtract.
• Hardest. Evaluate a definite integral of a power of x in a single step.

Not sure where to start? Take the ten-question placement test.