The actual *definition* of ‘integral’ is as a limit of sums, which
might easily be viewed as having to do with *area*. One of the
original issues integrals were intended to address was computation of
area.

First we need more notation. Suppose that we have a function
$f$ whose integral is another function $F$:
$$\int f(x)\;dx=F(x)+C$$
Let $a,b$ be two numbers. Then the **definite integral** of $f$ **with limits** $a,b$ is
$$\int_a^b f(x)\;dx=F(b)-F(a)$$
The left-hand side of this equality is just *notation* for the
definite integral. The use of the word ‘limit’ here has little to do
with our earlier use of the word, and means something more like
‘boundary’, just like it does in more ordinary English.

A similar notation is to write $$[g(x)]_a^b=g(b)-g(a)$$ for any function $g$. So we could also write $$\int_a^b f(x)\;dx=[F(x)]_a^b$$

For example, $$\int_0^5 x^2\;dx=\biggl[{x^3\over 3}\biggr]_0^5={5^3-0^3\over 3}={125\over 3}$$ As another example, $$\int_2^3 3x+1 \;dx=\biggl[{3x^2\over 2}+x\biggr]_2^3 =\biggl({3\cdot 3^2\over 2}+3\biggr)-\biggl({3\cdot 2^2\over 2}+2\biggr)={21\over 2}$$

All the other integrals we had done previously would be
called **indefinite integrals** since they didn't have ‘limits’
$a,b$. So a *definite* integral is just the difference of two
values of the function given by an *indefinite* integral. That is,
there is almost nothing new here except the idea of evaluating the
function that we get by integrating.

But now we *can* do something new: compute *areas*:

For example, if a function $f$ is *positive* on an
interval $[a,b]$, then
$$\int_a^b f(x)\;dx = \hbox{ area between graph and $x$-axis, between
$x=a$ and $x=b$}$$
It is important that the function be *positive*, or the result is
false.

For example, since $y=x^2$ is certainly always positive (or at least non-negative, which is really enough), the area ‘under the curve’ (and, implicitly, above the $x$-axis) between $x=0$ and $x=1$ is just $$\int_0^1 x^2\;dx=\biggl[{x^3\over 3}\biggr]_0^1={1^3-0^3\over 3}={1\over 3}$$

More generally, *the area below $y=f(x)$, above $y=g(x)$,
and between $x=a$ and $x=b$ is*
\begin{align*}\hbox{ area }&=\int_a^b f(x)-g(x) \;dx\\
&=\int_{\textit{left limit}}^{\textit{right limit}}
(\text{upper curve - lower curve}) \;dx
\end{align*}
It is important that $f(x)\ge g(x)$ throughout the interval $[a,b]$.

For example, the area below $y=e^x$ and above $y=x$, and between $x=0$ and $x=2$ is $$\int_0^2 e^x-x\;dx=\biggl[e^x-{x^2\over 2}\biggr]_0^2=(e^2-2)-(e^0-0)=e^2+1$$ since it really is true that $e^x\ge x$ on the interval $[0,2]$.

As a person might be wondering, in general it may be not so easy to tell whether the graph of one curve is above or below another. The procedure to examine the situation is as follows: given two functions $f,g$, to find the intervals where $f(x)\le g(x)$ and vice-versa:

- Find where the graphs cross by solving $f(x)=g(x)$ for $x$ to find the $x$-coordinates of the points of intersection.
- Between any two solutions $x_1,x_2$ of $f(x)=g(x)$ (and also to the
left and right of the left-most and right-most solutions!), plug in
*one*auxiliary point of your choosing to see which function is larger.

Of course, this procedure works for a similar reason that the *first derivative test* for local minima and maxima worked: we
implicitly assume that the $f$ and $g$ are *continuous*, so if the
graph of one is above the graph of the other, then the situation can't
*reverse* itself without the graphs actually *crossing*.

As an example, and as an example of a certain delicacy of
wording, consider the problem to *find the area between $y=x$ and
$y=x^2$ with $0\le x\le 2$*. To find where $y=x$ and $y=x^2$ *cross*, solve $x=x^2$: we find solutions $x=0,1$. In the present
problem we don't care what is happening to the left of $0$. Plugging
in the value $1/2$ as auxiliary point between $0$ and $1$, we get
${1\over 2}\ge ({1\over 2})^2$, so we see that in $[0,1]$ the curve
$y=x$ is the higher. To the right of $1$ we plug in the auxiliary
point $2$, obtaining $2^2\ge 2$, so the curve $y=x^2$ is higher
there.

Therefore, the area between the two curves has to be broken into two parts: $$\hbox{ area }=\int_0^1 (x-x^2)\; dx+\int_1^2 (x^2-x)\; dx$$ since we must always be integrating in the form $$\int_{\textit{left}}^{\textit{right}} \hbox{higher - lower}\;dx$$

In some cases the ‘side’ boundaries are redundant or only
*implied*. For example, the question might be to *find the
area between the curves $y=2-x$ and $y=x^2$*. What is implied here is
that these two curves themselves enclose one or more *finite*
pieces of area, without the need of any ‘side’ boundaries of the form
$x=a$. First, we need to see where the two curves intersect, by
solving $2-x=x^2$: the solutions are $x=-2,1$. So we *infer* that
we are supposed to find the area from $x=-2$ to $x=1$, and that the
two curves *close up* around this chunk of area without any need
of assistance from vertical lines $x=a$. We need to find which curve
is higher: plugging in the point $0$ between $-2$ and $1$, we see that
$y=2-x$ is higher. Thus, the desired integral is
$$\hbox{ area}=\int_{-2}^1 (2-x)-x^2 \; dx.$$

#### Exercises

- Find the area between the curves $y=x^2$ and $y=2x+3$.
- Find the area of the region bounded vertically by $y=x^2$ and $y=x+2$ and bounded horizontally by $x=-1$ and $x=3$.
- Find the area between the curves $y=x^2$ and $y=8+6x-x^2$.