Indefinite Integrals

The notation $\int f(x) dx = F(x)$ means that $F'(x) = f(x)$ on some (usually specified) domain of definition of $f(x)$.

Definition 2.2.1 (Anti-derivative)   We call $F(x)$ an anti-derivative of $f(x)$.

Proposition 2.2.2   Suppose $f$ is a continuous function on an interval $(a,b)$. Then any two antiderivatives differ by a constant.

Proof. If $F_1(x)$ and $F_2(x)$ are both antiderivatives of a function $f(x)$, then

$$(F_1(x) - F_2(x))' = F_1'(x) - F_2'(x) = f(x) - f(x) = 0.$$

Thus $F_1(x) - F_2(x) = c$ from some constant $c$ (since only constant functions have slope 0 everywhere). Thus $F_1(x) = F_2(x) + c$ as claimed. $\qedsymbol$

We thus often write

$$\int f(x) dx = F(x) + c,$$

where $c$ is an (unspecified fixed) constant.

Note that the proposition need not be true if $f$ is not defined on a whole interval. For example, $f(x) = 1/x$ is not defined at 0. For any pair of constants $c_1$, $c_2$, the function

$$F(x) = \begin{cases} \ln(\vert x\vert) + c_1 & x < 0,\\ \ln(x) + c_2 & x > 0, \end{cases}$$

satisfies $F'(x) = f(x)$ for all $x\neq 0$. We often still just write $\int 1/x = \ln(\vert x\vert)+c$ anyways, meaning that this formula is supposed to hold only on one of the intervals on which $1/x$ is defined (e.g., on $(-\infty,0)$ or $(0,\infty)$).

We pause to emphasize the notation difference between definite and indefinite integration.

$$\int_{a}^{b} f(x) dx =$$

$$\int f(x) dx =$$

One of the main goals of this course is to help you to get really good at computing $\int f(x)dx$ for various functions $f(x)$. It is useful to memorize a table of examples (see, e.g., page 406 of Stewart), since often the trick to integration is to relate a given integral to a known one. Integration is like solving a puzzle or playing a game, and often you win by moving into a position where you know how to defeat your opponent, e.g., relating your integral to integrals that you already know how to do. If you know how to do a basic collection of integrals, it will be easier for you to see how to get to a known integral from an unknown one.

Whenever you successfully compute $F(x) = \int f(x) dx$, then you've constructed a mathematical gadget that allows you to very quickly compute $\int_a^b f(x) dx$ for any $a,b$ (in the interval of definition of $f(x)$). The gadget is $F(b) - F(a)$. This is really powerful.