Video: Finding the Area Bounded by Curves of Two Rational Functions

The curves shown are 𝑦 = 1/π‘₯ and 𝑦 = 1/π‘₯Β². What is the area of the shaded region? Give an exact answer.

04:41

Video Transcript

The curves shown are 𝑦 equals one over π‘₯ and 𝑦 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 𝑦 equals 𝑓 of π‘₯, 𝑦 equals 𝑔 of π‘₯, and the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏 for continuous functions 𝑓 of 𝑔, when 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ for π‘₯ in the closed interval π‘Ž to 𝑏, is given by the definite integral evaluated between π‘Ž and 𝑏 of 𝑓 of π‘₯ minus 𝑔 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 can’t actually apply this definition. We can see, however, that if we split our region up a little more, we do achieve that requirement. I’ve 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. Let’s find the area of our first region, 𝑅 one. To do so, we’re 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 let’s 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, let’s substitute π‘₯ equals 0.5 into the equation 𝑦 equals one over π‘₯ squared. When we do, we get 𝑦 equals one over 0.5 squared, which is one over 0.25, which is four as required. So, the red line has the equation 𝑦 equals one over π‘₯ squared and the green line has equation 𝑦 equals one over π‘₯. And when evaluating the area of 𝑅 one, 𝑓 of π‘₯ is therefore one over π‘₯ squared and 𝑔 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 that’s 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‘₯.

We’re going to now evaluate this between π‘₯ equals 0.5 and π‘₯ equals one. That’s 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, I’ve 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. I’ve 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‘Ž to the 𝑏th power is equal to 𝑏 times the natural log of π‘Ž. We see that the exact area of the first region 𝑅 one is one minus the natural log of two.

Let’s clear some space and repeat this process for region two. This time, the green line is above the red line, so we’re going to let 𝑓 of π‘₯ be equal to one over π‘₯ and 𝑔 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. It’s 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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.