Indefinite Integrals

An antiderivative is simply a function whose derivative is the one you were given: you're running the derivative backwards. All we're really asking is "what did I differentiate to get this?" The notation looks new, but it's only a few pieces: $\int$ (read it as "the antiderivative of"), $f(x)$ (the function you were given), $dx$ (which says $x$ is the variable) and $+C$ (any constant, because constants disappear when you differentiate). It's a funny-looking way to ask a question you can already check.

So far in this course, we've been given a function and asked for its derivative. Now flip it around. If someone tells you the derivative is $2x$, what was the function? You probably thought of $x^2$ right away, because you know $(x^2)' = 2x$. That's an antiderivative.

Why $+C$? $x^2$, $x^2 + 5$ and $x^2 - 100$ all have derivative $2x$, since a constant's derivative is zero. And those are the only ones: by the Mean Value Theorem, two functions with the same derivative differ by a constant. So $x^2 + C$ describes every antiderivative of $2x$.

A common mistake. It's tempting to drop the $+C$, because $x^2$ alone looks like a perfectly good answer. It isn't the whole answer: $\int 2x\,dx$ asks for every antiderivative, and leaving off $C$ throws away all but one of them. It matters as soon as you need the specific one that fits a starting value.

Always check by differentiating. If $\frac{d}{dx}F(x) = f(x)$, you're right. This is the best habit in all of integration.

Properties (straight from the derivative rules):

• $\int k\,f(x)\,dx = k\int f(x)\,dx$
• $\int\big[f(x) \pm g(x)\big]dx = \int f(x)\,dx \pm \int g(x)\,dx$

There's no product or quotient rule for integrals, which is why integration needs a whole toolkit (substitution here, much more in Calculus II).

Antiderivatives run physics backwards: from acceleration recover velocity, from velocity recover position. The constant $C$ is the starting value you have to know.