Integration is a process for
computing bulk quantities by estimating and adding up smaller
pieces. This idea works in many different settings. In
this learning module we will develop the simplest application of
integration, namely finding the area under a curve.

Example 1: Find the area under the curve $y=2$ and above
the $x$-axis between $x=1$ and $x=4$.

Solution: The region is just a rectangle. The height of
the rectangle is $2$, and the width is $4-1=3$, so the area is 6.

Example 2: Find the approximate
area under the curve $y=f(x)$ and above the $x$ axis between $x=1$
and $x=4$, where $f(x) = 2 + 0.02 x$.

Solution: This is almost the same problem, except that the
height of the curve isn't constant. Even so, the height
doesn't change muchbetween
$x=1$ and $x=4$, so we can replace the curve $y=f(x)$ with the
horizontal line $y=f(1)$ to get an approximate total area of
$(4-1)f(1) = 3(2.02)=6.06$.

The answer in Example 2 is a little bit too small, an underestimate, since the curve goes
up from $y=2.02$ at $x=1$ to $y=2.08$ at $x=4$. If we wanted a
simple overestimate of the
area, we would use the value of $f(x)$ at the right endpoint
instead: $(4-1)f(4)=3(2.08)=6.24$. And if we wanted a more
accurate guess, we might use the midpoint
$f(2.5)=2.05$ and estimate the area as
$(4-1)f(2.5)=3(2.05)=6.15$. But whether you use the left
endpoint, the right endpoint, or the midpoint, you get roughly the
same answer, since the function just isn't changing very much
between $x=1$ and $x=4$.

So how do you find the area under a curve that is changing a lot? For
instance, how would you find the area under $y=x^2$ between $x=-1$
and $x=1$? The answer is to break the interval $[-1,1]$ up
into pieces, each so small that the function doesn't change much
in that piece. Then add up the pieces to get a good estimate
of the total. As with most of calculus, the exact answer is
then the limit of these
approximate answers.