7–58. Improper integrals Evaluate the following integrals or s...

Question

7–58. Improper integrals Evaluate the following integrals or state that they diverge.

53. ∫ (from 0 to 1) ln x dx

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate the improper integral of the natural log of X with respect to X between the bones of 0 and 3. A says it's 3 LN 3 minus 3. B says it's 3, C 3 LN 3, and the D says it diverges. Now let's first start by making sure we know that our integral really is improper. Now notice That LNX or function is continuous, OK. On the interval from open parentheses 0 to 3 close parentheses, but. At X equals 0, OK. But Add X equals 0. Our function diverges to negative infinity. So this really is an improper integral due to a vertical asymptote at the lower limit x equals 0. So since we know that, then we can rewrite this as a limit. In other words, We can rewrite the integral between 0 and 3 of LNX with respect to X. As the limit as a approaches 0 from the right between A and 3, that is the integral between A and the 3 of LN X with respect to X. Now we can go ahead and solve for an integral and then apply our limit. Now, we can integrate by using parts. I know integrating by parts. First, we can let you be equal to LNX. Which makes DU be equal to 1 divided by X with respect to X, OK? On the other hand, if we let D V equal DX, OK, then that means V would be equal to X. So no, applying our integral by parts. OK. Then, the integral of LNX with respect to X is going to be equal to XLNX minus X plus C. And in that case, This means then that our integral between A and 3 of LNX with respect to X would be equal to X LNX minus X between the bones of A and 3. And if we apply our upper and lower bones, then this will be 3 LN 3 minus 3 when we substitute 3 minus A LNA minus A when we substitute A. So now remember we'll be finding the limit of both these expressions. So taking the limit. As it approaches 0 from the right, OK. Then for the limit. But I should have actually just written that out. What for our limit A and then A minus A. Then we know that as A approaches 0 from the right, LNA approaches negative infinity, but A approaches 0, so ALNA will also approach 0. So by Lapita's rule on ALNA, the limit of our first term, ALNA as A approaches 0 from the right will be zero. We've confirmed that. And the limit as A approaches 0 of A will also be equal to 0. So this will be 0 minus 0, which equals 0. So really then. If this term if this limit will evaluate to zero, then. Our limit, if we take the limit of our other term as A approaches 0 of 3 LN 3 minus 3. Since all of these are constants, there's no A term here. This will still be 3 LN 3 minus 3. Therefore, this tells us then that our limit between 0 and 3 of LNX with respect to X will be equal to 3 LN3 minus 3. If we look back at our answer choices, that means that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

7–58. Improper integrals Evaluate the following integrals or state that they diverge.
53. ∫ (from 0 to 1) ln x dx

Key Concepts

Improper Integrals

Natural Logarithm Function Behavior near Zero

Integration by Parts

Watch next