Approximating Area Under a Curve
Graphically, we can consider a definite integral, such as
to be the area "under the curve", which might be better said as
the area that lies between the lines and the -axis
and the curve In order to find this area, we can begin
with a familiar geometric object: the rectangle. In this case, we
wish to find an area above the interval from to . In order to approximate
this, we can divide the interval into, say shorter intervals of
equal length. Let's call this length , and since they are
all the same length, we know that the length of each will be
However, in order to define an area, our rectangles require a height
as well as a width. Most often, calculus teachers will use the function's
value at the left or right endpoint for the height of each rectangle,
although we could also choose the minimum or maximum value of
on each interval, or perhaps the value at the midpoint of each interval.
Let's approximate this area first using left endpoints. Notice that
our leftmost interval is so the height at the left endpoint
is This means the area of our leftmost rectangle is
Continuing, the adjacent interval is Now, our left endpoint
is , and our area is
The next interval to the right is and as such the left
endpoint is so the area is
Finally, we have the rightmost rectangle, whose base is the interval This has as its left endpoint, so its area is
Adding these four rectangles up with sigma notation, we
can approximate the area under the curve as
Of course, we could also use right endpoints. In this case, we would
use the endpoints and 1 for each interval from
left to right to find
Note that in this case, one is an overestimate and one is an underestimate. This approximation through the area of rectangles is known as a Riemann sum.
Additional Examples with Fixed Numbers of Rectangles
Example 1. Approximate the area under the curve of
from to using rectangles and left endpoints.
Solution. Note that our -values range from to , so our
length is actually Thus each rectangle will have a base
This is our first step. This means our intervals from left to right
are and Choosing left endpoints,
Here is where the idea of "area under the curve" becomes clearer.
We actually have a signed area, where area below the -axis is
negative, while area above the -axis is positive.
Example 2. Approximate the area under the curve of
from to using rectangles and midpoints.
Solution. Here, our -values range from to so
As a result, our intervals from left to right are
and More importantly, our midpoints occur at
and respectively; this is where we will evaluate the height of
each rectangle. Thus
Defining the Integral as a Limit
Although associating the area under the curve with four rectangles
gives us a really rough approximation, there's no reason we can't
continue to divide (partition) the interval into smaller pieces, and
then get closer to the actual area. The graphic on the right shows
precisely the idea. We can keep making the base of each rectangle,
or smaller and smaller, and we'll get a better approximation.
More importantly, we can continue this idea as a limit, leading to
the following definition.
Definition. We define the Definite Integral of on
For an introductory course, we usually have
so each rectangle has exactly the same base.
Using the Definition to Evaluate a Definite Integral
Frequently, students will be asked questions such as: Using the definition
of the definite integral, find the area under the curve of the function on the interval using right endpoints.
Rather than using "easier" rules, such as the power rule and the
Fundamental Theorem of Calculus, this requires us to use the the definition
just listed. The first step is to set up our sum. We have
and is just an arbitrary natural (or counting) number. This tells
For a given , our leftmost interval would start at and be
of length This describes the interval
On the other hand, our next interval would start where the leftmost
stopped, or , and it's length would also be
This is the interval
If we call the leftmost interval then we would have
Similarly, our second interval would be
If we were to continue this rule, we would have that for any
we could write
This allows us to determine where to choose our height for each interval.
Since we are asked to use right endpoints, we would want
for for and finally
This allows us to build the sum. For an arbitrary we would have
Using this result, we now have
Now, we have several important sums explained on another page. These
are the sum of the first numbers, the sum of the first squares,
and the sum of the first cubes:
Moreover, we have some basic rules for summation. These rules that
make sense in simpler notation, such as
work the same way in Sigma notation, meaning
Before worrying about the limit as , when we
write both and the in the denominator are just constants, like
the in As a result we have
where we applied the rule for the first squares. Finally, we
can look at this as being approximately for large so the limit as is Thus
A Left Point of View
What would change if we approached the above integral through left
endpoints, instead of right? We would only be changing our value for For example, the leftmost interval is
so our left endpoint is On the other hand, our second interval
so our left endpoint is For the interval
our left endpoint is Let's apply the same process as the last section to
this value. for height. We then have
From here, we use the special sums again. This means that
Thus our choice of endpoints makes no difference in the resulting
A More Advanced Example
For most Riemann Sum problems in an integral calculus class,
will always be the same width, and we will need to use the special
sums to evaluate the limit. However, what can we do if we wish to
determine the value of
Using rectangles of the same width as shown in the earlier animation
would result in a very messy sum which contains a lot of square roots!
This makes finding the limit nearly impossible. Instead, we could
consider the inverse function to the square root, which is squaring.
Instead of choosing let's consider
For example, while In particular,
since we indexed the leftmost point as this means that
Each will be a different width, but either endpoint
would be a square, so taking will not leave a square root
in our sum.
Now we have all the pieces. Let's use right endpoints for the height
of each rectangle, so
You will NEVER see something like this in a first year calculus class,
but it is just a reminder that the definition includes the indexed
for a reason; the rectangles don't need to be all of the same width!