### Video Transcript

The integral from zero to five of one divided by the cube root of five minus π₯ with respect to π₯ is convergent. What does it converge to?

In this question, weβre given a definite integral and weβre told that this definite integral is convergent. We need to determine the value it converges to. This means we need to evaluate our definite integral, and the first thing we should always check is βis this an improper integral?β And remember, thereβs two things we need to check if an integral is improper. First, we need to check if either of the limits of integration are positive or negative β. And in this case, we can see this isnβt true. Both our upper and lower limit of integration are finite.

Next, we need to check if our integrand is continuous across the entire interval of integration. Since our upper limit of integration is five and the lower limit of integration is zero, we need to check if our integrand is continuous on the closed interval from zero to five. To do this, letβs take a closer look at our integrand. Itβs one divided by the cube root of five minus π₯. Using our laws of exponents, we could write this as five minus π₯ all raised to the power of negative one over three. So we can write our integrand as the composition between the linear function and the power function. And both of these are continuous. So our integrand is the composition of continuous functions and therefore is continuous across its entire domain.

So letβs find the domain of our integrand. We know we can take the cube root of any number; however, weβre not allowed to divide by zero. So the denominator of our integrand is not allowed to be equal to zero. And this will only happen when π₯ is equal to five. So our integrand is continuous for all values of π₯ except when π₯ is equal to five. And we can see π₯ is equal to five is the upper limit of integration. In other words, this is an improper integral. So to evaluate this integral, we need to recall how we evaluate improper integrals where our integrand is continuous everywhere on our interval of integration except for the upper limit of integration.

We recall if a function π is continuous for all values of π₯ greater than or equal to π and π₯ less than π and has a discontinuity when π₯ is equal to π, then the integral from π to π of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches π from the left of the integral from π to π‘ of π of π₯ with respect to π₯. And this is provided that this limit exists, which means it must be finite. And weβve shown this is exactly what we have for our integral. Our value of π, the lower limit of integration, is zero, and the upper limit of integration π is equal to five. And our integrand is continuous everywhere on this interval except when π is equal to five. So we can just apply this to the improper integral given to us in the question.

We get the integral from zero to five of one divided by the cube root of five minus π₯ with respect to π₯ is equal to the limit as π‘ approaches five from the left of the integral from zero to π‘ of one divided by the cube root of five minus π₯ with respect to π₯, provided this limit exists. And itβs worth pointing out something here, our values of π‘ are approaching five from the left. So our values of π‘ are less than five. Theyβre never equal to five. This means our integrand is now continuous across the entire interval of integration. So our integrand is now continuous across the interval of integration. In other words, this is now a proper integral, so we can use any of our tools for integration to help us evaluate this integral.

And to evaluate this integral, weβre going to use substitution. Weβre going to use the substitution π’ is equal to five minus π₯. So we want to use the substitution π’ is equal to five minus π₯. We need to differentiate both sides of this expression with respect to π₯. We get dπ’ by dπ₯ is equal to negative one. And of course, we know dπ’ by dπ₯ is not a fraction; however, when using integration by substitution, it can help to treat dπ’ by dπ₯ a little bit like a fraction. This gives us the equivalent statement in terms of differentials: dπ’ is equal to negative one dπ₯. And sometimes youβll just see this written as negative dπ₯.

Now remember, weβre using our substitution on a definite integral, so we need to find the new limits of integration. To do this, we need to substitute our values of π₯ into our expression for π’. Weβll start with the new upper limit of integration. We substitute π₯ is equal to π‘ into our expression for π’. We get π’ is equal to five minus π‘. Weβll then do the same for the lower limit of integration. We substitute π₯ is equal to zero into our expression for π’. We get π’ is equal to five minus zero which is, of course, just equal to five. Weβre now ready to apply our substitution π’ is equal to five minus π₯.

First, we just found the new lower limit of integration is five and the new upper limit of integration is five minus π‘. Next, in our integrand, weβll replace five minus π₯ with π’ giving us a new integrand of one over the cube root of π’. Finally, we can look at our equation in terms of differentials. We have dπ’ is equal to negative one dπ₯. If we divide both sides of this equation through by negative one, we get negative one dπ’ is equal to dπ₯. So we can replace dπ₯ with negative one dπ’. And again, itβs worth pointing out this is not exactly what is happening; however, it is a useful tool to think of it this way. And by doing this, we get the limit as π‘ approaches five from the left of the integral from five to five minus π‘ of one divided by the cube root of π’ times negative one with respect to π’.

And now we want to start evaluating this expression. First, we want to take the constant factor of negative one outside of our integral. But then we can see we can just take it outside of our limit as well. Doing this gives us the following expression. Next, we want to evaluate our integral. We can do this by using the power rule for integration. If we rewrite our integrand by using our laws of exponents, our integrand is π’ to the power of negative one over three. And now we can integrate this by using our power rule for integration. We want to add one to our exponent of π’, giving us a new exponent of two over three, and then divide by this new exponent of two over three. Doing this gives us the following limit we need to evaluate. In fact, we can simplify this further.

First, instead of dividing by two-thirds, we could multiply by the reciprocal of two-thirds. But this is a constant, so, in fact, we can just take this outside of our limit. Doing this, we get negative three over two multiplied by the limit as π‘ approaches five from the left of π’ to the power of two over three evaluated at the limit of integration π’ is equal to five and π’ is equal to five minus π‘. Now, weβre going to need to evaluate our antiderivative at the limits of integration. Doing this, we get negative three over two multiplied by the limit as π‘ approaches five from the left of five minus π‘ all raised to the power of two over three minus five raised to the power of two over three.

And now we want to evaluate this limit. And remember, the first thing we should always do when weβre asked to evaluate a limit of this form is see if we can use direct substitution. And in fact, we can in this case. We see the second term in this limit is a constant. And the first term in this limit is the composition between a linear function and a power function. These are continuous functions, so theyβre continuous across their entire domain. And we can see that π‘ is equal to five is in the domain of this function. So itβs continuous when π‘ is equal to five. So we can evaluate this limit by using direct substitution.

And this is where we see something interesting. When we substitute π‘ is equal to five, our first term will be zero minus zero raised to the power of two over three. So this is just equal to zero. So the first term inside of this limit is just equal to zero. This gives us negative three over two multiplied by negative five to the power of two over three. And we can then simplify this expression. The two negatives cancel, and this gives us our final answer: three over two multiplied by five to the power of two over three.

Therefore, we were able to show the integral from zero to five of one divided by the cube root of five minus π₯ with respect to π₯ converges to three over two multiplied by five to the power of two over three.