### Video Transcript

In this video, weβll learn how to
apply integration to find the area bounded by the curves of two or more
functions. By this stage, you should feel
confident in applying processes for integration to evaluate definite and indefinite
integrals. Weβll now look at how integration
can help us to find the area of the regions that lie between the graphs of two or
more functions.

Consider the region that lies
between the curve with equation π¦ equals π of π₯ and the π₯-axis which is bounded
by the vertical lines π₯ equals π and π₯ equals π. Iβve shaded this region pink. If π is a continuous function, we
know that we can evaluate the area of this region by integrating the function π of
π₯ with respect to π₯ between the limits of π and π.

Now, letβs add another curve to our
diagram. This time, the curve has the
equation π¦ equals π of π₯, where π is continuous and π of π₯ is less than or
equal to π of π₯ in the closed interval π to π. Once again, we can find the area of
the region between the curve π¦ equals π of π₯ and the π₯-axis and these two
vertical lines, now Iβve shaded this region in yellow, by evaluating the integral of
π of π₯ between the limits of π and π.

We can now see that if we subtract
the area between the curve π of π₯ and the π₯-axis from the area between the curve
of π of π₯ and the π₯-axis, that weβll be left with this region π΄ three. This is the region between the two
curves π¦ equals π of π₯ and π¦ equals π of π₯. We can, therefore, say that the
area π΄ three of the region bounded by π¦ equals π of π₯ and π¦ equals π of π₯ is
the integral of π of π₯ evaluated between π and π minus the integral of π of π₯
evaluated between π and π.

But we also know that the sum or
difference of the integral of two functions is equal to the integral of the sum or
difference of these functions. So, we can say that the area is
equal to the integral of π of π₯ minus π of π₯ with respect to π₯ evaluated
between π₯ equals π and π₯ equals π. This brings us to our first
definition. The area π΄ of the region bounded
by the curves π¦ equals π of π₯ and π¦ equals π of π₯, and the lines π₯ equals π
and π₯ equals π, where π and π are continuous and π of π₯ is greater than or
equal to π of π₯ in the closed interval π to π, is the definite integral of π of
π₯ minus π of π₯ evaluated between the limits of π and π.

Notice here that π of π₯ is
greater than or equal to π of π₯ for all π₯ in between and including π and π. Weβll need to watch carefully for
situations where this is not the case and apply some extra logic. For now though, weβll look at the
application of this formula.

Find the area of the region bounded
by the curves π¦ equals three π₯ squared minus five π₯ and π¦ equals negative five
π₯ squared.

Weβll recall that the area of the
region bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯, and the lines
π₯ equals π and π₯ equals π for continuous functions π and π, such that π of π₯
is greater than or equal to π of π₯ for all π₯ in the closed interval π to π, is
the definite integral between the limits of π and π of π of π₯ minus π of
π₯. Weβre, therefore, going to need to
define the functions π of π₯ and π of π₯ really carefully and, of course, the
values for π and π, ensuring that π of π₯ is greater than π of π₯ in the closed
interval π to π.

The lines π₯ equals π and π₯
equals π will mark the beginning and end of the region weβre interested in. So, what are the equations of these
lines? Theyβre the π₯-coordinates at the
points where the two graphs intercept. So, we can set the equations three
π₯ squared minus five π₯ and negative five π₯ squared equal to each other and solve
for π₯.

We begin by adding five π₯ squared
to both sides. And then, we factor the expression
on the left-hand side by taking out that factor of π₯. And we obtain π₯ times eight π₯
minus five to be equal to zero. We know that for this statement to
be true, either π₯ itself must be equal to zero or eight π₯ minus five must be equal
to zero. To solve this equation on the
right, we add five and then divide through by eight. And we obtain π₯ to be equal to
five-eighths.

So, we can see that the
π₯-coordinates of the points of intersection of our two curves are zero and
five-eighths. So, π is equal to zero and π is
equal to five-eighths. Now, weβre going to need to decide
which function is π of π₯ and which function is π of π₯. What we do next is sketch out the
graphs of π¦ equals three π₯ squared minus five π₯ and π¦ equals negative five π₯
squared. Weβre looking to establish which of
the curves is essentially on top.

We know that the graph of π¦ equals
three π₯ squared minus five π₯ is a U-shaped parabola. We can even factor the expression
three π₯ squared minus five π₯, set it equal to zero, and solve for π₯. And we see that it passes through
the π₯-axis at zero and five-thirds. So, it will look a little something
like this. The graph of π¦ equals negative
five π₯ squared is an inverted parabola which passes through the origin like
this. And so, we obtain the region
shaded.

We can now see that in the closed
interval of zero to five-eighths, the function thatβs on top, if you will, is the
function defined by π¦ equals negative five π₯ squared. So, we can say that π of π₯ is
equal to negative five π₯ squared. Meaning, π of π₯ is three π₯
squared minus five π₯. The area that weβre interested in
must, therefore, be given by the definite integral evaluated between zero and
five-eighths of negative five π₯ squared minus three π₯ squared minus five π₯ with
respect to π₯.

Distributing the parentheses, and
our integrand becomes negative eight π₯ squared plus five π₯. But wait a minute, we know that
when we evaluate areas below the π₯-axis, we end up with a funny result. We get a negative value. You might wish to pause the video
for a moment and consider what that means in this example. Did you work it out? We can see that our entire region
sits below the π₯-axis and weβre just working out the difference between the
areas. So, the negative results that we
would obtain from integrating each function individually will simply cancel each
other out. So, all thatβs left is to evaluate
this integral.

The integral of negative eight π₯
squared is negative eight π₯ cubed over three. And the integral of five π₯ is five
π₯ squared over two. We need to evaluate this between
zero and five-eighths, which is negative eight-thirds of five-eighths cubed plus
five over two times five-eighths squared minus zero. Thatβs 125 over 384 square
units. This question was fairly
straightforward as the curve of π¦ equals negative five π₯ squared was greater than
or equal to the curve of π¦ equals three π₯ squared minus five π₯ in the interval
weβre interested in.

Letβs now look at what we might do
if this wasnβt the case.

The curves shown are π¦ equals one
over π₯ and π¦ equals one over π₯ squared. What is the area of the shaded
region? Give an exact answer.

Remember, the area of a region
bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯, and the lines π₯
equals π and π₯ equals π for continuous functions π of π, when π of π₯ is
greater than or equal to π of π₯ for π₯ in the closed interval π to π, is given
by the definite integral evaluated between π and π of π of π₯ minus π of π₯.

Now, we do have a little bit of a
problem here. We can see quite clearly that the
region is bounded by the vertical lines π₯ equals 0.5 and π₯ equals two. But in the closed interval π₯ from
0.5 to two, we can see that one of our functions is not always greater than or equal
to the other. So, we canβt actually apply this
definition. We can see, however, that if we
split our region up a little more, we do achieve that requirement. Iβve added a third line at the
point where the two curves intersect. This has the equation π₯ equals
one.

In the closed interval 0.5 to one,
the values on the red line are always greater than or equal to those on the green
line. And in the closed interval π₯
between one and two, the reverse is true. So, all we do is split our region
up and then add the values at the end. Letβs find the area of our first
region, π
one. To do so, weβre going to need to
double check which line is which. We can probably deduce that the red
line is more likely to be one over π₯ squared. But letβs choose a coordinate pair
and substitute these values in just to be safe.

We can see that the curve passes
through the point with coordinates 0.5, 4. So, letβs substitute π₯ equals 0.5
into the equation π¦ equals one over π₯ squared. When we do, we get π¦ equals one
over 0.5 squared, which is one over 0.25, which is four as required. So, the red line has the equation
π¦ equals one over π₯ squared and the green line has equation π¦ equals one over
π₯. And when evaluating the area of π
one, π of π₯ is therefore one over π₯ squared and π of π₯ is equal to one over
π₯.

The area is, therefore, given by
the definite integral between the limits of 0.5 and one of one over π₯ squared minus
one over π₯. So, all thatβs left here is to
evaluate this integral. This is much easier to do if we
rewrite one over π₯ squared as π₯ to the power of negative two and then recall some
standard results. To integrate π₯ to the power of
negative two, we add one to the power and then divide by this new number. That gives us π₯ to the power of
negative one over negative one, which is negative one over π₯. The integral of one over π₯,
however, is the natural log of the absolute value of π₯. So, our integral is negative one
over π₯ minus the natural log of the absolute value of π₯.

Weβre going to now evaluate this
between π₯ equals 0.5 and π₯ equals one. Thatβs negative one over one minus
the natural log of one minus negative one over 0.5 minus the natural log of 0.5. And notice, Iβve lost the symbol
for the absolute value because one and 0.5 are already positive. The natural log of zero is one. Negative one over one is negative
one. And negative one over 0.5 is
two. Iβve also rewritten the natural log
of 0.5 as the natural log of a half and distributed the parentheses. And this simplifies to one plus the
natural log of one-half.

A really important skill, though,
is to be able to spot when we can further simplify a logarithmic term. If we rewrite the natural log of a
half as the natural log of two to the power of negative one. And then, use the fact that the
natural log of π to the πth power is equal to π times the natural log of π. We see that the exact area of the
first region π
one is one minus the natural log of two.

Letβs clear some space and repeat
this process for region two. This time, the green line is above
the red line, so weβre going to let π of π₯ be equal to one over π₯ and π of π₯ be
equal to one over π₯ squared. Our area is the definite integral
between one and two of one over π₯ minus one over π₯ squared which, when we
integrate, gives us the natural log of the absolute value of π₯ plus one over
π₯. Evaluating between the limits of
one and two, and we get the natural log of two plus a half minus the natural log of
one plus one, which is equal to the natural log of two minus a half.

We want to find the area of the
whole region, so we add these two values. Itβs one minus the natural log of
two plus the natural log of two minus one-half, which simplifies to one-half. The area of the shaded region is a
half square units. In this example, we saw that the
area formula can be applied to find the area between two curves where one curve is
above the other for part of the integration interval and the opposite in the second
part of the interval, as long as we remember to split the region up at the point
where the curves intersect.

Weβll now see how we can develop
this formula further to help us find the region bounded by three curves.

Find the area of the region bounded
by the curves π¦ equals four minus π₯ squared, π¦ equals negative π₯, and π¦ equals
the square root of π₯. Give your answer correct to one
decimal place.

Remember, for continuous functions
π and π, the area of the region bounded by the curves π¦ equals π of π₯, π¦
equals π of π₯, and the lines π₯ equals π and π₯ equals π, as long as π of π₯ is
greater than or equal to π of π₯ in the closed interval π to π, is given by the
integral evaluated between π and π of π of π₯ minus π of π₯. We are going to need to be a little
bit careful here, as we have three curves. So, letβs begin by sketching this
out and see what weβre dealing with.

The area enclosed between the three
curves looks a little something like this. Now, if weβre really clever, we can
actually use the definition we looked up before. We can split this region into the
region above the π₯-axis and the region below the π₯-axis. We then can split this up a little
bit further. We see that we have π
one, thatβs
the region between the π₯-axis and the curve π¦ equals root π₯ between π₯ equals
zero and π₯ equals π. Where π is the π₯-coordinate at
the point of intersection of the curve π¦ equals root π₯ and π¦ equals four minus π₯
squared.

We then have π
two. Thatβs the region between π¦ equals
four minus π₯ squared, π₯ equals π, and π₯ equals two. And the reason weβve chosen π₯
equals two as our upper limit is thatβs the π₯-value at the point at which the curve
crosses the π₯-axis. We can even split our three up into
two further regions to make life easier. But letβs deal first with the area
of π
one and π
two. We need to work out the value of
π.

We said itβs the π₯-coordinate at
the point of intersection of the two curves four minus π₯ squared and root π₯. So, we set these equal to each
other and solve for π₯. That gives us an π₯-value of 1.648,
correct to three decimal places. So, π is equal to 1.648. We can either do this by hand or
use our graphical calculators to evaluate each of these integrals. The area of π
one becomes 1.4104
and so on. And the area of π
two is 0.23326
and so on.

Letβs now consider the area of π
three. Itβs the integral between zero and
two of negative π₯ evaluated with respect to π₯. We need to be a little bit careful
here, since this is below the π₯-axis and therefore will yield a negative result on
integration. In fact, it gives us negative
two. So, we can say that the area is the
absolute value of this. Itβs two. And notice, we could have actually
used the formula for area of a triangle to work this area out.

Now, the area of π
four is the
definite integral between two and π of four minus π₯ squared minus negative π₯. And here, π is the π₯-coordinate
of the point of intersection of the lines π¦ equals negative π₯ and π¦ equals four
minus π₯ squared. We can once again set four minus π₯
squared equal to negative π₯ and solve for π₯. And we find, correct to three
decimal places, that they intersect at the point where π₯ equals 2.562. We type this into our calculator
and we find that the area of this region is 0.59106. We find the total of these four
values, which gives us 4.2347, which is 4.2 square units, correct to one decimal
place.

In this video, weβve seen that we
can use the formula area is equal to the definite integral between π and π of π
of π₯ minus π of π₯ with respect to π₯ for continuous functions, π and π as long
as π of π₯ is greater than or equal to π for all π₯ in the closed interval π to
π. We also saw that for more
complicated regions such as those bounded by three or more curves, those which
involve regions above and below the π₯-axis, or those where π of π₯ and π of π₯
switch, we might need to split this region up a little further.