### Video Transcript

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.