### Video Transcript

The curve shown is π¦ is equal to one divided by π₯. What is the area of the shaded region? Give an exact answer.

Weβre given a graph of the curve π¦ is equal to one over π₯. And thereβs a shaded region in this graph. We need to determine the area of the shaded region, and we need to give an exact answer. First, letβs determine what this region is. We can see that this is bounded by the line π₯ is equal to one and by the line π₯ is equal to one-third. We can also see that this region is bounded below by the π₯-axis. In other words, this entire region lies above the π₯-axis. Finally, this region is bounded above by our curve π¦ is equal to one over π₯.

To find the area of this curve, we need to recall what we mean by a definite integral. We know if a function π of π₯ is continuous for values of π₯ between π and π, then the area bounded by the curve π¦ is equal to π of π₯, the lines π₯ is equal to π and π₯ is equal to π is given by the definite integral from π to π of π of π₯ with respect to π₯ provided that π of π₯ is greater than or equal to zero for values of π₯ greater than or equal to π and less than or equal to π. In other words, we can find areas of regions bounded above by a curve by using integration.

And in fact we can see this is the case we have here. First, our function π of π₯ is equal to one over π₯. We can see our region is bounded by the lines π₯ is equal to one and π₯ is equal to one-third. So weβll set π equal to one and π equal to one-third. And of course, we already know this region is bounded by the π₯-axis. Finally, to use integration, we do need our function to be continuous on this interval. Luckily, one over π₯ is a rational function, so this will be continuous everywhere except where its denominator is equal to zero. So its only point of discontinuity is when π₯ is equal to zero. In particular, this means itβs continuous on this interval.

Therefore, we can find the area of the region given to us in the diagram by calculating the definite integral from one-third to one of one over π₯ with respect to π₯. To evaluate this integral, we need to recall the integral of the reciprocal function with respect to π₯ is equal to the natural logarithm of the absolute value of π₯ plus the constant of integration πΆ. Of course, in this case, weβre calculating a definite integral. So we donβt actually need our constant of integration. It will cancel in the working out. So we now have the natural logarithm of the absolute value of π₯ evaluated at the limits of integration π₯ is equal to one-third and π₯ is equal to one.

Now all thatβs left to do is evaluate this at the limits of integration. We get the natural logarithm of the absolute value of one minus the natural logarithm of the absolute value of one-third. And of course we can simplify this. First, the absolute value of one is just equal to one. But then the natural logarithm of one is just equal to zero. And of course the absolute value of one-third is just equal to one-third. So this is equal to negative the natural logarithm of one-third. And we could leave our answer like this. However, thereβs one more piece of simplification we can do.

We need to recall the power rule for logarithms. For the natural logarithm, this tells us π times the natural logarithm of π is equal to the natural logarithm of π raised to the power of π. So by doing this, instead of multiplying our entire natural logarithm by negative one, we can instead raise one-third to the power of negative one. So by using the power rule for logarithms, we now have the natural logarithm of one-third raised to the power of negative one. In other words, we need to take the reciprocal of one-third. But the reciprocal of one-third is just equal to three. So this simplifies to give us the natural logarithm of three.

Therefore, by using integration, we were able to calculate the area of the region given to us in the question. This was the region bounded by the curve π¦ is equal to one over π₯, the π₯-axis, the lines π₯ is equal to one and π₯ is equal to one-third. We found that this area was equal to the natural logarithm of three.