Question Video: Evaluating the Improper Integral of a Function with Infinite Limits of Integration Mathematics

Video Transcript

The integral from three to ∞ of one divided by 𝑥 minus two to the power of three over two with respect to 𝑥 is convergent. What does it converge to?

The question gives us a definite integral where one of the limits of our integral is ∞. We’re told that this integral is convergent, and we need to find what the value of this integral converges to. The first thing we need to do is set up a limit to find what this integral converges to. We’ll start by checking where our integrand is continuous.

The first thing we notice is our integrand is the composition of continuous functions. It’s a linear function raised to the power of three over two, and then we take the reciprocal. And we know the composition of continuous functions will be continuous on its entire domain. So, what is the domain of our integrand? Well, we can’t divide by zero. So, 𝑥 cannot be equal to two. And since we have a power of three over two, we’ll be taking the square root of 𝑥 minus two. So, 𝑥 minus two cannot be negative.

So, the domain of our integrand is all values of 𝑥 where 𝑥 minus two is greater than zero. And of course, we can just add two to both sides of this inequality. This gives us the domain of our integrand, and the set on which our integrand is continuous, is all real values of 𝑥 greater than two. This is useful because our integral is from three to ∞, our integrand is continuous for all values of 𝑥 greater than or equal to three. So, the only part of our integral which is improper is the upper limit of our integral being ∞.

Let’s now recall what we mean by an integral where the upper limit is equal to ∞. We say the integral from 𝑎 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is equal to the limit as 𝑡 approaches ∞ of the integral from 𝑎 to 𝑡 of 𝑓 of 𝑥 with respect to 𝑥 if this limit exists. And if this limit does not exist, then we say our integral is divergent. We’re told we’re dealing with a convergent integral, so this limit should exist in this case. So, let’s construct this limit for the integral given to us in the question.

The lower limit of our integral is three, so we’ll set 𝑎 equal to three, and we’ll set our function 𝑓 of 𝑥 to be our integrand. So, using our definition with 𝑎 equal to three and 𝑓 of 𝑥 as our integrand, we get the integral from three to ∞ of one divided by 𝑥 minus two to the power of three over two with respect to 𝑥 is equal to. The limit as 𝑡 approaches ∞ of the integral from three to 𝑡 of one divided by 𝑥 minus two to the power of three over two with respect to 𝑥 if this limit exists.

We now see we have a definite integral, which we need to evaluate. There’s a few different ways of evaluating this integral. We’ll do this by using our 𝑢 substitution. We’ll use 𝑢 is equal to 𝑥 minus two. Setting 𝑢 equal to 𝑥 minus two and then differentiating both sides with respect to 𝑥, we get d𝑢 by d𝑥 is equal to one. And remember, d𝑢 by d𝑥 is not a fraction. However, when we’re using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement d𝑢 is equal to d𝑥.

There’s one more thing we need to do before we use our 𝑢 substitution. We need to change the limits of are integral. Let’s start with the upper limit of our integral. That’s when 𝑥 is equal to 𝑡. We substitute this into our equation for 𝑢. This gives us 𝑢 is equal to 𝑡 minus two. Let’s now find the lower limit of our integral. We substitute 𝑥 is equal to three into our expression for 𝑢. We get 𝑢 is equal to three minus two, which is equal to one. So, by using our 𝑢 substitution, we’ve rewritten our definite integral as the integral from one to 𝑡 minus two of one divided by 𝑢 to the power of three over two with respect to 𝑢.

We can then rewrite our integrand by using our laws of exponents: one over 𝑢 to the power of three over two is the same as 𝑢 to the power of negative three over two. We can then integrate this by using our power rule for integration. We want to add one to our exponent of 𝑢 and then divide by this new exponent of 𝑢. This gives us the integral of 𝑢 to the power of negative three over two with respect to 𝑢 is equal to 𝑢 to the power of negative one-half divided by negative one-half plus the constant of integration 𝐶.

We can then multiply the numerator and the denominator of this by two and simplify by using our laws of exponents to get negative two divided by root 𝑢 plus 𝐶. So, applying this to evaluate our integral, we get the limit as 𝑡 approaches ∞ of negative two divided by root 𝑢 evaluated at the limits of our integral: 𝑢 is equal to one and 𝑢 is equal to 𝑡 minus two. Evaluating this at the limits of our integral, we get the limit as 𝑡 approaches ∞ of negative two divided by the square root of 𝑡 minus two plus two divided by root one.

We can evaluate the limit as 𝑡 approaches ∞ of both of these two terms separately. For our first term, the numerator remains constant. However, as 𝑡 is approaching ∞, our denominator is approaching ∞. So, our first term is approaching zero. And our second term is a constant; it doesn’t change as the value of 𝑡 changes. So, our limit evaluates to give us two divided by the square root of one, which we know is equal to two.

So, we’ve shown that our limit converges and it’s equal to two. And remember, this is the same as saying that our integral converges and our integral is equal to two. Therefore, we’ve shown the integral from three to ∞ of one divided by 𝑥 minus two to the power of three over two with respect to 𝑥 is convergent and it’s equal to two.