### 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.

We recall, the area π΄ of a region bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯ and the vertical lines π₯ equals π and π₯ equals π. Where π and π are continuous functions. Where π of π₯ is always greater than or equal to π of 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₯ with respect to 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 or equal to π of π₯ in that closed interval π to π.

The lines π₯ equals π and π₯ equals π mark the beginning and the end of the shaded region. By adding these vertical lines onto our diagram, we see that they have the equations π₯ equals 0.5 and π₯ equals one, respectively. And, we can therefore say that π is equal to 0.5 and π is equal to one in this example. In the closed interval, 0.5 to one, we see that the function thatβs on top, if you will, is the function defined by the red line. So, is this the function one over π₯ or one over π₯ squared? We can probably deduce that itβs more likely to be one over π₯ squared. But, letβs double check by choosing a coordinate pair and substituting these values in.

We can see that the red line passes through the point with coordinates 0.5, four. So, letβs substitute π₯ equals 0.5 into the equation π¦ equals one over π₯ squared. When we do, we get π¦ to be equal to one over 0.5 squared. Well, 0.5 squared is 0.25. And, one divided by 0.25 is four, as required. So, the red line is described by the equation π¦ equals one over π₯ squared. And, we can let π of π₯ be equal to one over π₯ squared, meaning π of π₯ is equal to one over π₯.

The area of our shaded region must therefore be given by the definite integral of one over π₯ squared minus one over π₯, evaluated between 0.5 and one. So, all thatβs left is to evaluate this definite integral. Now, this is actually slightly 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. So, thatβs π₯ to the power of negative one divided by negative one, which is negative one over π₯. The integral of one over π₯, however, is equal to the natural log of the absolute value of π₯.

So, the area is therefore equal to negative one over π₯ minus the natural log of the absolute value of π₯ evaluated between 0.5 and one. Weβll now substitute in those limits. And, we get negative one over one minus the natural log of one minus negative one over 0.5 minus the natural log of 0.5. Notice that since one and 0.5 are already positive, Iβve not worried about including the absolute value symbol in this part of our solution. The natural log of one is zero. So, this becomes negative one minus zero minus negative two minus the natural log of one-half.

Remember, weβre looking for an exact answer. So, weβre going to need to do something a little bit clever with our natural log of one-half. Weβll begin, though, by distributing the parentheses. And, we end up with negative one plus two plus the natural log of one-half, which is one plus the natural log of one-half. Itβs a really important skill to be able to spot when we can simplify a long rhythmic term. Here, we write a half as two to the power of negative one. And then, we use the fact that the natural log of π to the power of π is the same as π times the natural log of π. And, we obtain the shaded region to have an area of one minus the natural log of two square units.