In Exercises 35–68, use integration, the Direct Comparison Tes...

Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from -1 to 1 of (-x ln|x| dx)

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Determine whether the improper integral converges or diverges and find its value if it converges. The integral evaluated from 0 to 1 of the natural log of x d x. OK. So it appears for this particular problem we are given an improper integral and we're asked to determine if this particular integral converges or diverges and if it converges we're asked to figure out its value if it does converge. So now that we know what we're ultimately trying to solve for, let us note based on our prior knowledge, we need to be able to recall and recognize that the integrand has an improper behavior at x is equal to 0 since, since the natural log of x approaches negative infinity, so we need to interpret. The integral as if it is a limit. So we need to interpret the integral as a limit, meaning we need to take the integral, evaluate from 0 to 1 of the natural log of x DX is equal to the limit. As A approaches 0 from the right of the integral evaluated from A to 1 of the natural log of x dx. Fantastic. So now we need to evaluate our integral by recalling and using the integration by parts method, which states, or rather our formula for the integration by parts states the integral of uDV is equal to u multiplied by v minus the integral of VDU. So using our integration by parts formula and using the integration by parts method, let us state that you for this problem will be equal to the natural log of x. Then that will mean that duu by dx will be equal to 1 divided by x and that duu will be equal to dx divided by x. OK. And let us state that DV will be equal to DX, so that will mean that V will be simply equal to X. So now we need to substitute these values into our formula so we can go ahead and write that the integral evaluated from a to 1 of the natural log of x dx will be equal to x multiplied by the natural log of x evaluated from a to 1. Minus the integral evaluated from a to 1 of x multiplied by dx divided by x. Which will be equal to square brackets of x multiplied by the natural log of x minus x evaluated from a to 1, which will be equal to parentheses of 1 multiplied by the natural log of 1 minus 1 minus square brackets of a multiplied by the natural log of a minus a. Which will be all equal to -1 minus a multiplied by the natural log of a plus a. So now at this stage, we need to find the limit, so we need to take the limit as A approaches 0 from the right of the integral evaluated from A to 1 of the natural log of X D X will be equal to drum roll. The limit as a approaches 0 from the right of square brackets of -1 minus a multiplied by the natural log of a plus a. Which will be equal to the limit as a approaches 0 from the right of a multiplied by the natural log of a. Plus the limit, so it's gonna be plus the limit as A approaches 0 from the right. Of a Fantastic. So we now need to evaluate each limit. So we need to note that the limit as A approaches zero from the right of -1 will be equal to -1. And the limit as A approaches 0 from the right of A multiplied by the natural log of A. Will be equal to square brackets of 0 from the right multiplied by the natural log of 0 from the right. Which will be equal to square brackets of 0 multiplied by infinity, which will give us an indeterminate answer, and this will mean that it will require us to use algebraic manipulation to solve for this particular limit. So with that in mind. We need to take the limit as A approaches 0 from the right of a multiplied by the natural log of a. Which will be equal to the limit as a approaches 0 from the right of the natural log of a divided by 1 divided by a, which will be equal to square brackets of the natural log of 0 from the right divided by 1 divided by 0 from the right. Which will be equal to square brackets of infinity divided by infinity, which once again gives us an indeterminate answer. So now we need to recall and use Lahopatal's rule. So as we should recall, in order to apply Lahopatal's rule, we need to differentiate the numerator and denominator separately with respect to x. And after we apply Lahapatal's rule, if the expression still is in an indeterminate form, that means we will need to repeat the differentiation process again and again until a determinate form is achieved. So with that said, we need to take the limit. As a approaches 0 from the right of d by d a of the natural log of a divided by d by d a of 1 divided by a. Which will be equal to the limit as A approaches 0 from the right of drum roll, 1 divided by a divided by. -1 divided by. A to the power of 2. Which will be equal to the limit as A approaches 0 from the right of -8, which will be simply equal to 0. So that means that the limit as A approaches 0 from the right of A will be equal to 0. Thus that means we could safely write that the limit as a approaches 0 from the right of the integral evaluated from a to 1 of the natural log of x dx will be equal to the limit as a approaches 0 from the right. Of -1, fantastic. The limit -1 minus the limit. So once again, minus the limit as A approaches 0 from the right of A multiplied by the natural log of A. Plus, the limit as A approaches 0 from the right of A will be all equal to -1 minus 0 + 0, which will be simply equal to -1. So therefore our improper integral converges. Hooray, it converges, and it converges to -1, and that's it. That is our final answer. Hooray, we did it. So to recap what we've learned, we could safely say when we evaluated our improper integral, we can safely say once again that our final answer is converges to -1. And that's it, that is our final answer. Thank you so much for watching, hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.∫ from -1 to 1 of (-x ln|x| dx)

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In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.
∫ from -1 to 1 of (-x ln|x| dx)