### Video Transcript

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.