### Video Transcript

Determine the area of the plane region bounded by the curve π¦ is equal to eight divided by π₯ to the fourth power and the lines π₯ is equal to one, π₯ is equal to eight, and π¦ is equal to zero rounded to the nearest hundredth.

The question wants us to determine the area of a region which is bounded by the curve π¦ is equal to eight divided by π₯ to the fourth power and the lines π₯ is equal to one, π₯ is equal to eight, and π¦ is equal to zero. It wants us to give our answer rounded to the nearest hundredth.

We recall that we can calculate the region above the π₯-axis bounded by the curve π¦ is equal to π of π₯ and the lines π₯ is equal to π and π₯ is equal to π when π is less than π by using the integral from π to π of π of π₯ with respect to π₯. In fact, we could also use this to calculate the area below the π₯-axis bounded by the curve π¦ is equal to π of π₯ and the lines π₯ is equal to π and π₯ is equal to π, where π is less than π. We do this by taking negative the integral from π to π of π of π₯ with respect to π₯.

So now that we have a method of calculating the area above or below the π₯-axis, letβs sketch a graph of the region given to us in the question. Weβll start by drawing our vertical lines π₯ is equal to one and π₯ is equal to π and a horizontal line π¦ is equal to zero, which is coincident with the π₯-axis. Next, weβll sketch the graph of π¦ is equal to eight divided by π₯ to the fourth power. We know the graph of one divided by π₯ to the fourth power is a reciprocal graph. And π₯ is raised to an even power. So it would lie entirely above the π₯-axis. Then, multiplying this by π is a vertical stretch of factor eight. So it will still stay above the π₯-axis.

So we now have a sketch of our region, which we will label π
. And since we know the curve π¦ is equal to eight divided by π₯ to the fourth power stays above the π₯-axis, we know that this region is above the π₯-axis. Itβs worth noticing at this point that this formula for calculating the area wonβt work if we try to integrate over a region where our curve is discontinuous. However, our curve of π¦ is equal to eight divided by π₯ to the fourth power is a rational function. The only times it can possibly be discontinuous is when its denominator is equal to zero. And in this case, it is discontinuous when π₯ is equal to zero. However, in our case, we have the bounds for our integral as π₯ is equal to one and π₯ is equal to eight. So our curve is continuous over this region.

So weβre now ready to try and calculate the area of our region. We have that this area is equal to the integral from one to eight of eight divided by π₯ to the fourth power with respect to π₯. We can rewrite eight divided by π₯ to the fourth power as eight multiplied by π₯ to the power of negative four. Now, we can calculate this integral by recalling for constants π and π, where π is not equal to negative one, to integrate π multiplied by π₯ to the πth power with respect to π₯, we add one to our exponent and divide by this new exponent. Then, we add our constant of integration, π.

Adding one to our exponent gives us negative three. And then, we divide ππ₯ to the negative three by negative three. And since weβre calculating a definite integral, we donβt need to add a constant of integration. Evaluating this to our limits of π₯ is equal to one and π₯ is equal to eight gives us eight multiplied by eight to the power of negative three over negative three minus eight multiplied by one to the power of negative three over negative three. Which to the nearest one hundredth is 2.66.

Therefore, weβve shown that the region bounded by the curve π¦ is equal to eight divided by π₯ to the fourth power and the lines π₯ is equal to one, π₯ is equal to eight, and π¦ is equal to zero rounded to the nearest one hundredth is 2.66.