Video Transcript
Consider the region in the first
quadrant enclosed by the curves 𝑦 is equal to four over 𝑥, 𝑦 is equal to 𝑥, and
𝑦 is equal to 𝑥 over four. Find the area of this region.
In this question, we need to
determine the area of a region enclosed by three given curves, the curve 𝑦 is equal
to four over 𝑥 and the lines 𝑦 is equal to 𝑥 and 𝑦 is equal to 𝑥 over four. And whenever we’re asked to find
the area of a region enclosed by curves, it’s always a good idea to sketch the
curves given. This will then allow us to see the
region we need to find the area of. Let’s start by sketching the two
straight lines, 𝑦 is equal to 𝑥 and 𝑦 is equal to 𝑥 over four. Both of these are given in
slope–intercept form, and we can see that their 𝑦-intercepts are zero. And the slope of 𝑦 is equal to 𝑥
is one, and the slope of 𝑦 is equal to 𝑥 over four is one-quarter. So we can sketch both of the
straight lines as shown.
Now, on the same coordinate axis,
we need to sketch the curve 𝑦 is equal to four over 𝑥. And we can see that this is the
reciprocal function one over 𝑥 multiplied by four. So this will have the shape of a
reciprocal curve. Adding this to our diagram gives us
a shape which looks like the following. And remember, the area of the
region we need to determine is only in the first quadrant. So we can add this region onto our
diagram as shown.
To help us evaluate this integral,
we need to recall one of our integral results. If we have two integrable
functions, 𝑓 of 𝑥 and 𝑔 of 𝑥, where 𝑓 of 𝑥 is greater than or equal to 𝑔 of
𝑥 on the closed interval from 𝑎 to 𝑏, then the area of the region enclosed by the
curves 𝑦 is equal to 𝑓 of 𝑥 and 𝑦 is equal to 𝑔 of 𝑥 and the vertical lines 𝑥
is equal to 𝑎 and 𝑥 is equal to 𝑏 is given by the integral from 𝑎 to 𝑏 of 𝑓 of
𝑥 minus 𝑔 of 𝑥 with respect to 𝑥.
We want to apply this to find the
area of the region in our diagram. However, there’s one small
problem. We can see that we have three
functions, so we’re going to need to split our region in two. We can split our region into two
parts: one where the upper bound is 𝑦 is equal to 𝑥 and one where the upper bound
is 𝑦 is equal to four over 𝑥. We can then find the area of both
parts of this region by using our integral result, where it’s important to note we
do know that all of the functions are integrable in the first quadrant. That means 𝑥 is greater than zero
since all three functions are continuous for positive values of 𝑥.
Therefore, let’s find the area of
each region separately. Let’s start with the left part of
the region. The upper part of this region is
the line 𝑦 is equal to 𝑥, and the lower part of this region is 𝑦 is equal to 𝑥
over four. These will be our functions 𝑓 of
𝑥 and 𝑔 of 𝑥. Next, to apply this result, we need
to find the values of 𝑎 and 𝑏. We can see the value of 𝑎 directly
from the diagram. The two lines intersect when 𝑥 is
equal to zero at the origin, so our value of 𝑎 is zero. However, to find the value of 𝑏,
we’re going to need to find the 𝑥-coordinate of the point of intersection between
the line 𝑦 is equal to 𝑥 and the curve 𝑦 is equal to four over 𝑥. And we can do this by setting the
two functions equal to each other. We need to solve 𝑥 is equal to
four over 𝑥.
We can solve this equation by first
multiplying through by 𝑥. We get 𝑥 squared is equal to
four. And now we take the square root of
both sides of the equation, where we note we’ll get a positive and a negative
root. 𝑥 is either equal to two or
negative two. And we can see in the diagram what
each of these solutions represent. 𝑥 is negative two is the leftmost
point of intersection between the line and the curve. And 𝑥 is equal to positive two is
the rightmost intersection. Since we’re only working in the
first quadrant, we’re only interested in the rightmost value. So our value of 𝑏 is two.
We can now substitute all of this
into our integral result to find the area of the leftmost region. It’s the integral from zero to two
of 𝑥 minus 𝑥 over four with respect to 𝑥. We could now evaluate this
expression. However, we also need to add on the
area of the second region. So let’s find an expression for
this area. This time, the region is bounded
above by the curve 𝑦 is equal to four over 𝑥 and below by the line 𝑦 is equal to
𝑥 over four. We’ve already found the value of
𝑎. In this case, our value of 𝑎 is
two. We now need to find the value of
𝑏, and this will be the point of intersection between the curve 𝑦 is equal to four
over 𝑥 and the line 𝑦 is equal to 𝑥 over four. And we can find this value in a
similar way we did before. We need to solve the functions
equal to each other. We need to solve the equation 𝑥
over four is equal to four over 𝑥.
First, we’ll multiply both sides of
the equation through by four 𝑥. This gives us that 𝑥 squared is
equal to 16. Then we take the square root of
both sides of the equation, where we get a positive and a negative root. 𝑥 is equal to four or negative
four. So there are two points of
intersection between this line and curve: one when 𝑥 is equal to negative four and
one when 𝑥 is equal to positive four. And we’re working in the first
quadrant, so our values of 𝑥 are positive. Our value of 𝑏 is four. We can now substitute all of this
into our integral result to find the area of the rightmost part of the region. It’s the integral from two to four
or four over 𝑥 minus 𝑥 over four with respect to 𝑥. And now, the sum of these two areas
will be the area of the region we’re asked to determine.
All that’s left to do now is
evaluate both of these integrals. First, we can simplify the leftmost
integrand to be three 𝑥 over four. We can now evaluate both of these
integrals by recalling two integral results. First, the power rule for
integration tells us for any real constants 𝑎 and 𝑛, where 𝑛 is not equal to
negative one, the integral of 𝑎𝑥 to the 𝑛th power with respect to 𝑥 is equal to
𝑎 times 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus the constant of
integration 𝐶. We add one to the exponent of 𝑥
and then divide by this new exponent. Second, we can also recall for any
real constant 𝑎, the integral of 𝑎 over 𝑥 with respect to 𝑥 is 𝑎 times the
natural logarithm of the absolute value of 𝑥 plus the constant of integration
𝐶.
We can use these two results to
evaluate both of these integrals. Let’s start with our first
integral. We need to add one to the exponent
of 𝑥 and divide by this new exponent. Adding one to the exponent of 𝑥
gives us a new exponent of two. And then we divide by two to get
three 𝑥 squared divided by four times two, which is eight. In our second integral, we
integrate four over 𝑥 with respect to 𝑥 to get four times the natural logarithm of
the absolute value of 𝑥. And to integrate negative 𝑥 over
four, we add one to the exponent of 𝑥 to get 𝑥 squared and divide by this new
exponent. We get negative 𝑥 squared over
eight.
All that’s left to do now is
evaluate the antiderivatives at the limits of integration. In our first antiderivative, we can
see the lower limit of integration is zero. And when we substitute this into
our antiderivative, we get zero. So we only need to substitute in
the upper limit of integration. We get three times two squared over
eight. The same is not true in the second
case. We need to substitute the upper
limit of integration into our antiderivative. And then we need to subtract from
this the lower limit of integration substituted into our antiderivative. This gives us the following
expression for the area.
Now, all that’s left to do is
evaluate this expression. First, three times two squared over
eight is 12 over eight. Next, negative four squared over
eight is negative 16 over eight. And third, negative two squared
over eight is negative four over eight. But remember, we’re subtracting
this term, so we need to add four over eight. We can then see that 12 over eight
minus 16 over eight plus four over eight is equal to zero. So we’re left with the two terms
four times the natural logarithm of the absolute value of four minus four times the
natural logarithm of the absolute value of two. And we can simplify this
further. We know that the absolute value of
a positive number doesn’t change its value, and both four and two are positive. So we can just remove the absolute
value symbol.
Next, we’re going to rewrite the
natural logarithm of four by using the power rule for logarithms. To do this, we need to rewrite four
as two squared. This then allows us to take the
exponent outside of the logarithm as a multiplier by using the aforementioned rule
of logarithms. This gives us eight times the
natural logarithm of two minus four times the natural logarithm of two, which
simplifies to give us four times the natural logarithm of two, which is our final
answer.
Therefore, we were able to show the
area of the region in the first quadrant enclosed by the curves 𝑦 is equal to four
over 𝑥, 𝑦 is equal to 𝑥, and 𝑦 is equal to 𝑥 over four is four times the
natural logarithm of two.
