### Video Transcript

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.