As noted in the first section of this section there are two
kinds of integrals and to this point we’ve looked at indefinite integrals. It is now time to start thinking about the
second kind of integral : Definite Integrals.
However, before we do that we’re going to take a look at the Area
Problem. The area problem is to definite
integrals what the tangent and rate of change problems are to derivatives.
The area problem will give us one of the interpretations of
a definite integral and it will lead us to the definition of the definite
integral.
To start off we are going to assume that we’ve got a
function that is positive on some interval [a,b].
What we want to do is determine the area of the region between the function
and the xaxis.
It’s probably easiest to see how we do this with an
example. So let’s determine the area
between on [0,2].
In other words, we want to determine the area of the shaded region
below.
Now, at this point, we can’t do this exactly. However, we can estimate the area. We will estimate the area by dividing up the
interval into n subintervals each of
width,
Then in each interval we can form a rectangle whose height
is given by the function value at a specific point in the interval. We can then find the area of each of these
rectangles, add them up and this will be an estimate of the area.
It’s probably easier to see this with a sketch of the
situation. So, let’s divide up the interval
into 4 subintervals and use the function value at the right endpoint of each
interval to define the height of the rectangle.
This gives,
Note that by choosing the height as we did each of the
rectangles will over estimate the area since each rectangle takes in more area
than the graph each time. Now let’s
estimate the area. First, the width of
each of the rectangles is . The height of each rectangle is determined by
the function value at the right endpoint and so the height of each rectangle is
nothing more that the function value at the right endpoint. Here is the estimated area.
Of course taking the rectangle heights to be the function
value at the right endpoint is not our only option. We could have taken the rectangle heights to
be the function value at the left endpoint.
Using the left endpoints as the heights of the rectangles will give the
following graph and estimated area.
In this case we can see that the estimation will be an
underestimation since each rectangle misses some of the area each time.
There is one more common point for getting the heights of
the rectangles that is often more accurate.
Instead of using the right or left endpoints of each sub interval we
could take the midpoint of each subinterval as the height of each
rectangle. Here is the graph for this
case.
So, it looks like each rectangle will over and under
estimate the area. This means that the
approximation this time should be much better than the previous two choices of
points. Here is the estimation for this
case.
We’ve now got three estimates. For comparison’s sake the exact area is
So, both the right and left endpoint estimation did not do
all that great of a job at the estimation.
The midpoint estimation however did quite well.
Be careful to not draw any conclusion about how choosing
each of the points will affect our estimation.
In this case, because we are working with an increasing function
choosing the right endpoints will overestimate and choosing left endpoint will
underestimate.
If we were to work with a decreasing function we would get
the opposite results. For decreasing
functions the right endpoints will underestimate and the left endpoints will
overestimate.
Also, if we had a function that both increased and decreased
in the interval we would, in all likelihood, not even be able to determine if
we would get an overestimation or underestimation.
Now, let’s suppose that we want a better estimation, because
none of the estimations above really did all that great of a job at estimating
the area. We could try to find a
different point to use for the height of each rectangle but that would be cumbersome
and there wouldn’t be any guarantee that the estimation would in fact be
better. Also, we would like a method for
getting better approximations that would work for any function we would chose
to work with and if we just pick new points that may not work for other
functions.
The easiest way to get a better approximation is to take
more rectangles (i.e. increase n).
Let’s double the number of rectangles that we used and see what
happens. Here are the graphs showing the
eight rectangles and the estimations for each of the three choices for
rectangle heights that we used above.
Here are the area estimations for each of these cases.
So, increasing the number of rectangles did improve the
accuracy of the estimation as we’d guessed that it would.
Let’s work a slightly more complicated example.
Example 1 Estimate
the area between and the xaxis
using subintervals and all three cases above for
the heights of each rectangle.
Solution
First, let’s get the graph to make sure that the function
is positive.
So, the graph is positive and the width of each
subinterval will be,
This means that the endpoints of the subintervals are,
Let’s first look at using the right endpoints for the
function height. Here is the graph for
this case.
Notice, that unlike the first area we looked at, the
choosing the right endpoints here will both over and underestimate the area
depending on where we are on the curve.
This will often be the case with a more general curve that the one we
initially looked at. The area
estimation using the right endpoints of each interval for the rectangle
height is,
Now let’s take a look at left endpoints for the function
height. Here is the graph.
The area estimation using the left endpoints of each
interval for the rectangle height is,
Finally, let’s take a look at the midpoints for the
heights of each rectangle. Here is the
graph,
The area estimation using the midpoint is then,
For comparison purposes the exact area is,

So, again the midpoint did a better job than the other
two. While this will be the case more
often than not, it won’t always be the case and so don’t expect this to always
happen.
Now, let’s move on to the general case. Let’s start out with on [a,b]
and we’ll divide the interval into n
subintervals each of length,
Note that the subintervals don’t have to be equal length,
but it will make our work significantly easier.
The endpoints of each subinterval are,
Next in each interval,
we choose a point . These points will define the height of the
rectangle in each subinterval. Note as
well that these points do not have to occur at the same point in each
subinterval.
Here is a sketch of this situation.
The area under the curve on the given interval is then
approximately,
We will use summation
notation or sigma notation at
this point to simplify up our notation a little. If you need a refresher on summation notation
check out the section devoted to this in
the Extras chapter.
Using summation notation the area estimation is,
The summation in the above equation is called a Riemann Sum.
To get a better estimation we will take n larger and larger. In fact,
if we let n go out to infinity we
will get the exact area. In other words,
Before leaving this section let’s address one more
issue. To this point we’ve required the
function to be positive in our work.
Many functions are not positive however.
Consider the case of on [0,2]. If we use and the midpoints for the rectangle height we
get the following graph,
In this case let’s notice that the function lies completely
below the xaxis and hence is always
negative. If we ignore the fact that the
function is always negative and use the same ideas above to estimate the area
between the graph and the xaxis we get,
Our answer is negative as we might have expected given that
all the function evaluations are negative.
So, using the technique in this section it looks like if the
function is above the xaxis we will
get a positive area and if the function is below the xaxis we will get a negative area.
Now, what about a function that is both positive and negative in the
interval? For example, on [0,2].
Using and midpoints the graph is,
Some of the rectangles are below the xaxis and so will give negative areas while some are above the xaxis and will give positive
areas. Since more rectangles are below
the xaxis than above it looks like
we should probably get a negative area estimation for this case. In fact that is correct. Here the area estimation for this case.
In cases where the function is both above and below the xaxis the technique given in the
section will give the net area
between the function and the xaxis
with areas below the xaxis negative
and areas above the xaxis
positive. So, if the net area is
negative then there is more area under the xaxis
than above while a positive net area will mean that more of the area is above
the xaxis.