In this section we are going to look at finding the area between
two curves. There are actually two cases
that we are going to be looking at.
In the first case we want to determine the area between 
and on the interval [a,b]. We are also going to
assume that . Take a look at the following sketch to get an
idea of what we’re initially going to look at.
In the Area and Volume
Formulas section of the Extras chapter we derived the following formula for
the area in this case.
The second case is almost identical to the first case. Here we are going to determine the area
between and on the interval [c,d] with .
In this case the formula is,
Now (1) and (2)
are perfectly serviceable formulas, however, it is sometimes easy to forget
that these always require the first function to be the larger of the two
functions. So, instead of these formulas
we will instead use the following “word” formulas to make sure that we remember
that the area is always the “larger” function minus the “smaller”
function.
In the first case we will use,
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(3)
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In the second case we will use,
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(4)
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Using these formulas will always force us to think about
what is going on with each problem and to make sure that we’ve got the correct
order of functions when we go to use the formula.
Let’s work an example.
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Example 1 Determine
the area of the region enclosed by and .
Solution
First of all, just what do we mean by “area enclosed
by”. This means that the region we’re
interested in must have one of the two curves on every boundary of the
region. So, here is a graph of the two
functions with the enclosed region shaded.
Note that we don’t take any part of the region to the
right of the intersection point of these two graphs. In this region there is no boundary on the
right side and so is not part of the enclosed area. Remember that one of the given functions
must be on the each boundary of the enclosed region.
Also from this graph it’s clear that the upper function
will be dependent on the range of x’s
that we use. Because of this you
should always sketch of a graph of the region. Without a sketch it’s often easy to mistake
which of the two functions is the larger.
In this case most would probably say that is the upper function and they would be
right for the vast majority of the x’s. However, in this case it is the lower of
the two functions.
The limits of integration for this will be the
intersection points of the two curves.
In this case it’s pretty easy to see that they will intersect at and so these are the limits of integration.
So, the integral that we’ll need to compute to find the
area is,
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Before moving on to the next example, there are a couple of
important things to note.
First, in almost all of these problems a graph is pretty
much required. Often the bounding
region, which will give the limits of integration, is difficult to determine
without a graph.
Also, it can often be difficult to determine which of the
functions is the upper function and which is the lower function without a
graph. This is especially true in cases
like the last example where the answer to that question actually depended upon
the range of x’s that we were using.
Finally, unlike the area under a curve that we looked at in
the previous chapter the area between two curves will always be positive. If we get a negative number or zero we can be
sure that we’ve made a mistake somewhere and will need to go back and find it.
Note as well that sometimes instead of saying region
enclosed by we will say region bounded by.
They mean the same thing.
Let’s work some more examples.
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Example 2 Determine
the area of the region bounded by ,
,
,
and the y-axis.
Solution
In this case the last two pieces of information, and the y-axis,
tell us the right and left boundaries of the region. Also, recall that the y-axis is given by the line . Here is the graph with the enclosed region
shaded in.
Here, unlike the first example, the two curves don’t meet.
Instead we rely on two vertical lines
to bound the left and right sides of the region as we noted above
Here is the integral that will give the area.
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Example 3 Determine
the area of the region bounded by and .
Solution
In this case the intersection points (which we’ll need
eventually) are not going to be easily identified from the graph so let’s go
ahead and get them now. Note that for
most of these problems you’ll not be able to accurately identify the
intersection points from the graph and so you’ll need to be able to determine
them by hand. In this case we can get
the intersection points by setting the two equations equal.
So it looks like the two curves will intersect at and . If we need them we can get the y values corresponding to each of
these by plugging the values back into either of the equations. We’ll leave it to you to verify that the
coordinates of the two intersection points on the graph are (-1,12) and
(3,28).
Note as well that if you aren’t good at graphing knowing
the intersection points can help in at least getting the graph started. Here is a graph of the region.
With the graph we can now identify the upper and lower
function and so we can now find the enclosed area.
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Be careful with parenthesis in these problems. One of the more common mistakes students make
with these problems is to neglect parenthesis on the second term.
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Example 4 Determine
the area of the region bounded by ,
,
and
Solution
So, the functions used in this problem are identical to
the functions from the first problem.
The difference is that we’ve extended the bounded region out from the
intersection points. Since these are
the same functions we used in the previous example we won’t bother finding
the intersection points again.
Here is a graph of this region.
Okay, we have a small problem here. Our formula requires that one function
always be the upper function and the other function always be the lower
function and we clearly do not have that here. However, this actually isn’t the problem
that it might at first appear to be.
There are three regions in which one function is always the upper
function and the other is always the lower function. So, all that we need to do is find the area
of each of the three regions, which we can do, and then add them all up.
Here is the area.
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Example 5 Determine
the area of the region enclosed by ,
,
,
and the y-axis.
Solution
First let’s get a graph of the region.
So, we have another situation where we will need to do two
integrals to get the area. The
intersection point will be where
in the interval.
We’ll leave it to you to verify that this will be . The area is then,
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We will need to be careful with this next example.
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Example 6 Determine
the area of the region enclosed by and .
Solution
Don’t let the first equation get you upset. We will have to deal with these kinds of
equations occasionally so we’ll need to get used to dealing with them.
As always, it will help if we have the intersection points
for the two curves. In this case we’ll
get the intersection points by solving the second equation for x and then setting them equal. Here is that work,
So, it looks like the two curves will intersect at and or if we need the full coordinates they will
be : (-1,-2) and (5,4).
Here is a sketch of the two curves.
Now, we will have a serious problem at this point if we
aren’t careful. To this point we’ve
been using an upper function and a lower function. To do that here notice that there are
actually two portions of the region that will have different lower
functions. In the range [-2,-1] the
parabola is actually both the upper and the lower function.
To use the formula that we’ve been using to this point we
need to solve the parabola for y. This gives,
where the “+” gives the upper portion of the parabola and
the “-” gives the lower portion.
Here is a sketch of the complete area with each region
shaded that we’d need if we were going to use the first formula.
The integrals for the area would then be,
While these integrals aren’t terribly difficult they are
more difficult than they need to be.
Recall that there is another formula for determining the
area. It is,
and in our case we do have one function that is always on
the left and the other is always on the right. So, in this case this is definitely the way
to go. Note that we will need to
rewrite the equation of the line since it will need to be in the form but that is easy enough to do. Here is the graph for using this formula.
The area is,
This is the same that we got using the first formula and
this was definitely easier than the first method.
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So, in this last example we’ve seen a case where we could
use either formula to find the area.
However, the second was definitely easier.
Students often come into a calculus class with the idea that
the only easy way to work with functions is to use them in the form . However, as we’ve seen in this previous
example there are definitely times when it will be easier to work with
functions in the form . In fact, there are going to be occasions when
this will be the only way in which a problem can be worked so make sure that
you can deal with functions in this form.
Let’s take a look at one more example to make sure we can
deal with functions in this form.
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Example 7 Determine
the area of the region bounded by and .
Solution
First, we will need intersection points.
The intersection points are and . Here is a sketch of the region.
This is definitely a region where the second area formula
will be easier. If we used the first
formula there would be three different regions that we’d have to look at.
The area in this case is,
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