Question Video: Evaluating the Improper Integral of a Discontinuous Function Involving a Root Function Mathematics • Higher Education

The integral ∫_(βˆ’2) ^(14) 1/∜(π‘₯ + 2) dπ‘₯ is convergent. What does it converge to?

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Video Transcript

The integral from negative two to 14 of one divided by the fourth root of π‘₯ plus two with respect to π‘₯ is convergent. What does it converge to?

In this question, we’re given a definite integral and we’re told that this definite integral is convergent. We need to determine what value does this definite integral converge to. And we should always check if our integral is improper. First, we should check if either of the limits of integration are positive or negative ∞. And in this case, we can see that neither of our limits of integration are positive or negative ∞. Our lower limit of integration is negative two, and the upper limit of integration is 14.

Next, we should check that our integrand is continuous on the interval of integration. In this case, we can see our integrand one divided by the fourth root of π‘₯ plus two is the composition of continuous functions. And the composition of continuous functions is always continuous across its entire domain. So, we just need to find the domain of our integrand. This comes in two steps. First, we know we’re not allowed to divide by zero. And second, we know we can’t take the fourth root of a negative number.

And of course, this just means that π‘₯ plus two is not allowed to be less than or equal to zero. Otherwise, our integrand is not defined. Either we’ll be dividing by zero or taking the fourth root of a negative number. And we’ll rearrange this inequality. We’re going to subtract two from both sides to get the new inequality π‘₯ is less than or equal to negative two. And this is where we can see that our integral is improper. Our integrand is not continuous at the lower limit of integration. And this is because our integrand is not even defined at this point.

Therefore, because this integral is improper, we’re going to need to recall how we evaluate the integral of an improper integral. We recall if a function 𝑓 of π‘₯ is continuous for all values of π‘₯ greater than π‘Ž and all values of π‘₯ less than or equal to 𝑏 and it has a discontinuity when π‘₯ is equal to π‘Ž, then the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is equal to the limit as 𝑑 approaches π‘Ž from the right of the integral from 𝑑 to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ as long as this limit exists and is finite.

Essentially, this gives us a method of turning an improper integral into the limit of a proper integral. And this is in fact guaranteed because, remember, the only discontinuous point of 𝑓 of π‘₯ on the closed interval from π‘Ž to 𝑏 should be at π‘Ž. So, because 𝑑 is approaching π‘Ž from the right, our value of 𝑑 is never equal to π‘Ž. So, the new interval of integration will be the closed interval from 𝑑 to 𝑏 and 𝑓 of π‘₯ must be continuous on this interval.

So, we want to use this to evaluate the improper integral given to us in the question. First, recall our integrand is the composition of continuous functions, so it must be continuous across its entire domain. In particular, this means it will be continuous for all values of π‘₯ greater than negative two and all values of π‘₯ less than or equal to 14. And of course, we’ve already shown it’s not continuous when π‘₯ is equal to negative two because it’s not even defined at this point. Therefore, we’ll set our value of π‘Ž equal to negative two and 𝑏 equal to 14.

We can then update our definition for the improper integral with π‘Ž equal to negative two and 𝑏 equal to 14. And of course, we know our integrand 𝑓 of π‘₯ is continuous on this interval and not continuous when π‘₯ is equal to negative two. So, we can try and use our rules for improper integrals to evaluate this integral. This gives us the integral from negative two to 14 of one divided by the fourth root of π‘₯ plus two with respect to π‘₯ is equal to the limit as 𝑑 approaches negative two from the right of the integral from 𝑑 to 14 of one divided by the fourth root of π‘₯ plus two with respect to π‘₯ as long as this limit exists and is finite.

It’s worth reiterating at this point 𝑑 is never equal to negative two, so our integral is now a proper integral. We can use any of our rules of integration to help us evaluate this integral. In particular, to help us evaluate this integral, we’ll want to use a substitution. We’re going to use 𝑒 is equal to π‘₯ plus two. So, we want to use the substitution 𝑒 is equal to π‘₯ plus two. We’ll start by differentiating both sides of this equation with respect to π‘₯. This gives us d𝑒 by dπ‘₯ is equal to one. And remember, d𝑒 by dπ‘₯ is not a fraction. However, when we’re doing integration by substitution, it can help us to treat it a little bit like a fraction.

This gives us the equivalent statement in terms of differentials, d𝑒 is equal to dπ‘₯. But remember, we’re using a 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 the equation 𝑒 is equal to π‘₯ plus two. We’ll start by finding the new upper limit of integration. We substitute π‘₯ is equal to 14. This gives us 𝑒 is equal to 14 plus two, which is of course just equal to 16.

And we do exactly the same thing for our lower limit of integration. We substitute π‘₯ is equal to 𝑑, giving us that 𝑒 is equal to 𝑑 plus two. We’re now ready to use our substitution. First, the new lower limit of integration is 𝑑 plus two and the new upper limit of integration is 16. Next, in our integrand, we’ll replace π‘₯ plus two with 𝑒. Finally, we need to remember by using our differentials, dπ‘₯ is equal to d𝑒. So, we now need to determine the limit as 𝑑 approaches negative two from the right of the integral from 𝑑 plus two to 16 of one divided by the fourth root of 𝑒 with respect to 𝑒.

And in fact, we can just evaluate this integral directly. To do this, we’re going to use our laws of exponents to rewrite our integrand as 𝑒 to the power of negative one-quarter. Now, we can just evaluate this by using the power rule for integration. We want to add one to our exponent of 𝑒 and then divide by this new exponent. Adding one to our exponent of negative one-quarter gives us a new exponent of three-quarters. And then, we need to divide by this new exponent of three-quarters. This gives us the following expression.

And we can simplify this. Instead of dividing by the fraction three-quarters, we can instead multiply it by the reciprocal, which is four over three. This gives us the limit as 𝑑 approaches two from the right of four 𝑒 to the power three over four divided by three evaluated at limits of integration 𝑑 plus two and 16.

The next thing we need to do is evaluate this at the limits of integration. Doing this, we get the limit as 𝑑 approaches negative two from the right of four times 16 to the power of three over four divided by three minus four times 𝑑 plus two all raised to the power of three over four all divided by three. And this is a very complicated-looking limit. However, we can actually evaluate this directly.

First, our entire first term is just a constant. Next, our second term is a continuous function. It’s the composition between a linear function and a power function. So, we can just evaluate this by substituting 𝑑 is equal to negative two into this expression. And if we substitute 𝑑 is equal to negative two into this, we see we get a factor of zero since negative two plus two is zero. So, the limit as 𝑑 approaches negative two of our second term is equal to zero.

Therefore, all we need to do is evaluate four times 16 to the power of three over four divided by three. And if we calculate this expression, we see it’s equal to 32 divided by three. Therefore, we’ve shown that this limit exists and it’s finite. And because it exists and it’s finite, it must be equal to our improper integral. Therefore, we were able to show the integral from negative two to 14 of one divided by the fourth root of π‘₯ plus two with respect to π‘₯ converges to 32 divided by three.

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