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We review differentiation and integration.

### Antiderivatives

As a summary, one of the important questions of differential calculus is “Given a function, what is its derivative?” There is an important related question in integral calculus that requires “undoing” the process of differentiation; that is, “Given $f'(x)$, what are the possibilities for $f(x)$?” This is made precise by the following definition:

There is an important, but subtle way this definition is phrased; note that we used the phrase “an antiderivative” and not “the antiderivative”. To expound:

Which of the following are antiderivatives of $2x$?
$x^2+5$ $x^2-3$ $2x^2+3$ $2-x^2$

Notice that there are several choices for antiderivatives of $x^2$. So, how could these antiderivatives differ? The answer is a consequence of the Mean Value Theorem, and we state the result below:

This theorem guarantees that the antiderivatives of a function $f(x)$ are the same up to an additive constant, which allows us to introduce the following notation:

Many of the helpful rules that allow us to find derivatives of more complicated functions are not easily reversible. We do have the following:

Find $\int \left (4x^2 + \dfrac {1}{4+x^2} \right )\d x$.

Unfortunately, there is no simple rule that allows us to compute antiderivatives of general products, quotients, or compositions of arbitrary functions as there were with differentiation.

Consider $f(x) = \cos (\sqrt {x})$. Which of the following is $\int f(x) \d x$?
$\sin (\sqrt {x}) +C$ $\dfrac {1}{\sqrt {x}}\sin (\sqrt {x}) +C$ $-\dfrac {1}{2 \sqrt {x}} \sin (\sqrt {x})+C$ $2 \cos (\sqrt {x}) + 2 \sqrt {x} \sin (\sqrt {x}) +C$ None of these

The following is a question that employs the same logic, but is phrased a bit differently:

### Definite Integrals

For continuous functions, the Fundamental Theorem of Calculus provides the link between the process of antidifferentiation and finding certain areas. Recall that for functions $f(x)$ continuous on a closed interval $[a,b]$, the symbol $\int _a^b f(x) \d x$ denotes the net area bounded by $y=f(x)$ and the $x$-axis between $x=a$ and $x=b$. By “net area”, recall that area above the $x$-axis is considered positive, and area below it is considered negative.

Computing this area initially had to be done by setting up Riemann sums and finding limits of them, but thankfully, there is a much more efficient way to do this:

As a final remark, note that the Fundamental Theorem of Calculus comes with assumptions. It can only be used if the integrand is continuous on the closed interval of integration and that interval is finite.