Question Video: Determining Whether the Improper Integral of a Rational Function Is Convergent or Divergent Mathematics

Video Transcript

Determine whether the integral from zero to five of 𝑤 divided by 𝑤 minus two with respect to 𝑤 is convergent or divergent.

In this question, we’re given a definite integral, and we’re asked to determine if this definite integral is convergent or divergent. And the first thing we should always check if we’re asked to evaluate a definite integral is, is our integral improper? And to check this, we need to check two things. First, we need to check whether either of our limits of integration are positive or negative ∞. And for this integral, we can see this is not the case. Both of our limits of integration are finite. So, our interval of integration is the closed interval from zero to five.

The next thing we need to check is, is our integrand continuous on the interval of integration? And to answer this question, we need to notice something about our integrand. It’s a rational function, and rational functions are continuous across their entire domain. And we know the only time a rational function won’t be defined is if we divide by zero. So, we just solve this denominator is equal to zero. This happens when 𝑤 is equal to two. So, our integrand is continuous for all real values of 𝑤 except when 𝑤 is equal to two. And this is where we have our problem. Two is in the interval of integration. So, this is an improper integral, and we’re going to need to evaluate it by using our rules for improper integrals.

In fact, there’s a few different ways of doing this. We’re going to use the following method. If we have a function 𝑓 which is continuous on a closed interval from 𝑎 to 𝑏 except at one value of 𝑐 where 𝑐 is between 𝑎 and 𝑏, then the integral from 𝑎 to 𝑏 of 𝑓 of 𝑤 with respect to 𝑤 is equal to the limit as 𝑠 approaches 𝑐 from the left of the integral from 𝑎 to 𝑠 of 𝑓 of 𝑤 with respect to 𝑤 plus the limit as 𝑡 approaches 𝑐 from the right of the integral from 𝑡 to 𝑏 of 𝑓 of 𝑤 with respect to 𝑤. And this is very complicated looking. However, all we’re doing is saying the top integrand is discontinuous at the point 𝑐.

So, by using our properties of definite integrals, we want to split our integral at the point 𝑐. But then this would give us two improper integrals. So, we need to evaluate each of these by using our rules for improper integrals. This gives us the following two limits. And there’s something worth pointing out here. We can only say that our limit is convergent and equal to a value if both of these limits are convergent. If either of these two limits are divergent, then our entire integral is divergent. We now want to apply this to the integral given to us in the question.

First, the limits of integration at zero and five. So, we’ll set our value of 𝑎 equal to zero and 𝑏 equal to five. Next, we’ve shown there’s only one point of discontinuity when 𝑤 is equal to two. So, we’ll set our value of 𝑐 equal to two. So, by substituting these values in and using our integrand, we have the integral from zero to five of 𝑤 divided by 𝑤 minus two with respect to 𝑤 is equal to the limit as 𝑠 approaches two from the left of the integral from zero to 𝑠 of 𝑤 divided by 𝑤 minus two with respect to 𝑤 plus the limit as 𝑡 approaches two from the right of the integral from 𝑡 to five of 𝑤 divided by 𝑤 minus two with respect to 𝑤. And of course, this is only true if both of these limits are convergent.

And there’s something worth pointing out about both of these limits. In our first limit, 𝑠 is approaching two from the left, and 𝑡 is approaching two from the right. So, our values of 𝑠 are always less than two, and our values of 𝑡 are always bigger than two. In particular, this means that 𝑠 is never equal to two, and 𝑡 is also never equal to two. So, now, both of our integrands are continuous across the entire interval of integration. So, we can use any of our tools to help us evaluate this integral. There’s actually lots of different methods we could use to evaluate this integral. For example, we could use the substitution 𝑢 is equal to 𝑤 minus two.

However, there’s actually an easier method in this case. Either by using inspection or by using algebraic division, we can show that 𝑤 divided by 𝑤 minus two is equal to one plus two divided by 𝑤 minus two. And this is a much easier expression to integrate. So, we’ll replace both of our integrands with one plus two divided by 𝑤 minus two. We’re now almost ready to evaluate this integral. However, there’s one more thing we can notice. We need to recall the following integral rule. The integral of 𝑔 prime of 𝑤 divided by 𝑔 of 𝑤 with respect to 𝑤 is equal to the natural logarithm of the absolute value of 𝑔 of 𝑤 plus the constant of integration capital 𝐶.

And we can see this is almost exactly what we have in two divided by 𝑤 minus two. The derivative of our denominator is the constant one. So, our numerator is a constant multiple of the derivative of our denominator. So, by taking out the factor of two, integrating just this expression will give us two times the natural logarithm of 𝑤 minus two where we add a constant of integration capital 𝐶. However, we don’t need to do this because we’re using a definite integral. We’re now ready to start evaluating this integral. In our first limit, the integral of one with respect to 𝑤 is 𝑤. Next, as we’ve already discussed, the integral of two divided by 𝑤 minus two is two times the natural logarithm of the absolute value of 𝑤 minus two.

Finally, we need to evaluate this at the limits of integration zero and 𝑠. We can then do exactly the same to evaluate the integral in our second limit. This gives us the following expression. The next thing we need to do is evaluate both of these expressions at the limits of integration. Evaluating our first expression at the limit of integration, we get the limit as 𝑠 approaches two from the left of 𝑠 plus two times the natural logarithm of the absolute value of 𝑠 minus two minus zero minus two times the natural logarithm of the absolute value of zero minus two. And we can simplify this expression. First, subtracting zero doesn’t change the value. Next, the absolute value of zero minus two is just equal to two. So, we’ve simplified our limit to give us the following expression.

We’ll now do the same with our second limit. We need to evaluate this at the limits of integration. Doing this, we get the limit as 𝑡 approaches two from the right of five plus two times the natural logarithm of the absolute value of five minus two minus 𝑡 minus two times the natural logarithm of the absolute value of 𝑡 minus two. And we can simplify this slightly. The absolute value of five minus two is equal to three. This means we were able to simplify our second limit into the following expression.

Now, remember for our integral to be convergent, both of these limits must be convergent. So, let’s check if this is true. We’ll start with our first limit. We see 𝑠 is approaching two from the left. And we can do this term by term. We see the limit as 𝑠 approaches two from the left of 𝑠 is just equal to two. And in our third term, we can see it’s a constant, so it’s not changing as the value of 𝑠 changes.

However, we see we have a problem with our second term, two times the natural logarithm of the absolute value of 𝑠 minus two. As 𝑠 approaches two from the left of this expression, we can see that it’s approaching negative ∞. And therefore, our first limit evaluates to give us negative ∞. But remember, when a limit is equal to negative ∞, this also means that it’s a divergent limit. Therefore, because one of these limits was not convergent, we can conclude our original integral must be divergent.

Therefore, we were able to show the integral from zero to five of 𝑤 divided by 𝑤 minus two with respect to 𝑤 is divergent.