Video Transcript
The integral from two to ∞ of 𝑦 times 𝑒 to the power of negative three 𝑦 with respect to 𝑦 is convergent. What does it converge to?
The question is asking us to evaluate a definite integral where the upper limit is ∞. In fact, we’re told that this integral is convergent. We need to find what this integral converges to. Since the upper limit of this integral is ∞, we know this is an improper integral. So we know we’re going to need to use our rules of improper integrals. The first thing we should do when we’re asked to evaluate a definite integral is check the continuity of our integrand. We do this because we want to check for any discontinuities on our interval of integration.
In our case, our integrand is the product of 𝑦 and 𝑒 to the power of negative three 𝑦. We know that 𝑦 is the identity function and 𝑒 to the power of negative three 𝑦 is an exponential function. Both of these functions are continuous for all real values of 𝑦. And we know the product of continuous functions is itself continuous. This means our integrand is continuous for all real values of 𝑦. In particular, this means our integrand is continuous on our interval of integration, all values of 𝑦 greater than or equal to two. So the only part of this definite integral which makes it improper is the upper limit being ∞.
Let’s now recall one of our rules for improper integrals. We know if 𝑓 of 𝑦 is a continuous function for all values of 𝑦 greater than or equal to two. Then, we can say the integral from 𝑎 to ∞ of 𝑓 of 𝑦 with respect to 𝑦 is equal to the limit as 𝑡 approaches ∞ of the integral from 𝑎 to 𝑡 of 𝑓 of 𝑦 with respect to 𝑦. And of course, this is only if this limit exists. If this limit does not exist, then we say that our integral is divergent. We’ve already shown this applies to the integral given to us in the question. So let’s use this rule on this integral. This gives us the integral from two to ∞ of 𝑦 times 𝑒 to the power of negative three 𝑦 with respect to 𝑦 is equal to. The limit as 𝑡 approaches ∞ of the integral from two to 𝑡 of 𝑦 times 𝑒 to the power of negative three 𝑦 with respect to 𝑦, assuming this limit exists.
Now, the upper limit of our definite integral is some finite value of 𝑡. And our integrand is continuous over this entire closed interval, so we can use any of our integral rules to evaluate this definite integral. There are several different ways we could evaluate this integral. We see that our integrand is the product of two functions. So we’ll try and do this by using integration by parts. We recall integration by parts tells us the integral from 𝑎 to 𝑏 of 𝑢 times 𝑣 prime with respect to 𝑦 is equal to 𝑢 times 𝑣 evaluated at the limits of our integral 𝑦 is equal to 𝑎 and 𝑦 is equal to 𝑏 minus the integral from 𝑎 to 𝑏 of 𝑢 prime 𝑣 with respect to 𝑦.
We could use the LIATE method to tell us which of our two factors to use as our function 𝑢. And since there are no logarithmic or inverse trigonometric functions, this tells us we should use an algebraic function, which we see is 𝑦. We could’ve also done this by noticing differentiating 𝑦 makes it simpler. So we’ll choose 𝑢 to be 𝑦 and 𝑣 prime to be 𝑒 to the power of negative three 𝑦. So we now have 𝑢 is equal to 𝑦 and 𝑣 prime is 𝑒 to the power of negative three 𝑦. We need to find expressions for 𝑢 prime and 𝑣. Well, to find 𝑢 prime, we just need to differentiate 𝑦 with respect to 𝑦. This is a linear function, so it’s just the coefficient of 𝑦, which is equal to one.
To find our expression for 𝑣, we need to integrate 𝑒 to the power of negative three 𝑦 with respect to 𝑦. To do this, we need to divide by the coefficient of 𝑦 in our exponent. This gives us 𝑣 is equal to negative one-third 𝑒 to the power of negative three 𝑦. We’re now ready to rewrite our definite integral by using integration by parts. We need to substitute our expressions for 𝑢, 𝑢 prime, 𝑣, and 𝑣 prime into our formula for integration by parts. This gives us the limit as 𝑡 approaches ∞ of 𝑦 times negative one-third 𝑒 to the power of negative three 𝑦 evaluated at the limits of our integral 𝑦 is equal to two and 𝑦 is equal to 𝑡. Minus the definite integral from two to 𝑡 of one times negative one-third 𝑒 to the power of negative three 𝑦 with respect to 𝑦.
We can then simplify this expression. To start, multiplying by one inside of our integrand does not change the value of our integral. We can then see that this is an integral which we know how to evaluate. We need to divide by the coefficient of 𝑦 in our exponent. Before we do this, let’s start by evaluating the first term inside of our limit at the limits of integration. This gives us 𝑡 times negative one-third 𝑒 to the power of negative three 𝑡 minus two times negative one-third times 𝑒 to the power of negative three times two. Now, we want to subtract our integral. And remember, we can evaluate this by dividing by the coefficient of 𝑦 in our exponent. The coefficient of 𝑦 in our exponent is equal to negative three. So we need to divide negative one-third by negative three. This gives us one-ninth.
We can then simplify this expression slightly. We’ll write the first term inside of our limit as negative 𝑡 divided by three 𝑒 to the power of three 𝑡 and the second term inside of our limit as two divided by three 𝑒 to the sixth power. This then gives us the following expression. We can simplify this even further. We need to evaluate the third term inside of our limit at the limits of integration. Doing this and then multiplying through by negative one, we get negative one-ninth 𝑒 to the power of negative three 𝑡 plus one-ninth 𝑒 to the power of negative three times two. And of course, negative three times two is equal to negative six.
Finally, we’ll use our laws of exponents to simplify this even further. This gives us the limit as 𝑡 approaches ∞ of negative 𝑡 divided by three 𝑒 to the power of three 𝑡 plus two divided by three 𝑒 to the sixth power. Minus one over nine 𝑒 to the power of three 𝑡 plus one over nine 𝑒 to the sixth power. We’re now almost ready to evaluate this limit. There’s just one last piece of simplification we can do. We know that two over three 𝑒 to the sixth power plus one over nine 𝑒 to the sixth power is equal to seven over nine 𝑒 to the sixth power. So we’ll combine these into one term. We’re now ready to evaluate this limit term by term.
Let’s start with our first term, negative 𝑡 divided by three 𝑒 to the power of three 𝑡. Our limit is as 𝑡 is approaching ∞. If we try to evaluate our limit directly, we’ll get negative ∞ divided by ∞. This is an indeterminant form, so we’re going to need to evaluate this limit in a different way. Since negative 𝑡 and three to the power of three 𝑡 are both continuous for all values of 𝑡, we’ll do this by using L’Hôpital’s rule.
Now, by using L’Hôpital’s rule, we get the limit of our first term is equal to the limit as 𝑡 approaches ∞ of negative 𝑡 prime divided by three 𝑒 to the power of three 𝑡 prime. If this limit exists or it’s equal to positive or negative ∞. We can evaluate both the derivative in our numerator and the derivative in our denominator. This gives us the limit as 𝑡 approaches ∞ of negative one divided by nine 𝑒 to the power of three 𝑡. And this time, as 𝑡 approaches ∞, our numerator remains constant. However, our denominator is unbounded. This means this limit evaluates to give us zero. So the limit of our first term is equal to zero by using L’Hôpital’s rule.
In fact, this also tells us the limit of our third term is also equal to zero. And finally, our third term is a constant. It doesn’t change as the value of 𝑡 changes. So the limit as 𝑡 approaches ∞ of this term is just equal to itself. We’ll write this as seven 𝑒 to the power of negative six divided by nine. And because this limit exists, this means our integral is convergent. In fact, this means we’ve shown the integral from two to ∞ of 𝑦 times 𝑒 to the power of negative three 𝑦 with respect to 𝑦 is equal to seven 𝑒 to the power of negative six divided by nine.
