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

Question

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

44. ∫ (from 0 to ln 3) eʸ/(eʸ-1)⁷ᐟ³ dy

Accepted Answer

Welcome back, everyone. In this problem, evaluate the improper integral or state that it diverges. The integral between 0 and the natural log of 4 of E to the divided by e Y minus 1 to 11/4 with respect to Y. A says it's negative4/7, B4/7, C 0, and D says that it diverges. Now, for starters, let's make sure we establish that this really is an improper integra. Now, notice that. If we were to examine the behavior of the integrand at the end points, OK, then at Y equals 0. Then e 0 would be equal to 1, and that implies that ey minus 1 would approach 0. Now, if E to the Y minus 1 is approaching 0, that means our denominator will be 0, and then our expression will be undefined because we cannot divide by 0. So this means this really is an improper integral at the lower limit. So in that case, we can rewrite it as a limit and then solve for the integra. In other words, we can then say that our integral of 0 integral between 0 and the natural log of 4 of E to the Y divided by E Y minus 1 to the 11/4 with respect to Y. is equal to the limit as a approaches 0 from the right of the integral between A and the natural log of 4 of e to the Y divided by e to the Y minus 1. To the 11th words with respect to why. Now we can evaluate our integral and put it back into our limit. Now let's evaluate it using substitution. So let you be equal to E to the Y minus 1. And that means the derivative of U with respect to Y will be equal to E Y, which means that DU will be equal to E Y D Y. Now, when Y equals A, OK, then U is going to be equal to A E A minus 1 and this value will approach 0 from the right as A. Approaches 0 from the right, OK. Next, For Y equals the natural log of 4. That means you would be equal to E to the natural log of 4 minus 1. But based on what we know about logarithms, we could rewrite this as 4 multiplied by LNE, which would just be 4, and 4 minus 1 equals 3. So now that we have our limits, you and the DU in terms of Y, then we can say. That our expression, the limit as a approaches 0 from the right of the given integral. I'll just right here. Yeah. It is not going to be equal to the limit as a small value epsilon approaches 0 from the right of the integral between epsilon and 3 of you to the power of negative 11/4 with respect to you. No, notice that. This is a standard PE integral with P equals 11/4, which is greater than 1, so it converges improperly near zero. So let's compute our integral, apply our limits, or apply our bones, and then see if we can solve for the limit. So now this means we're finding the limit as epsilon approaches 0 of our integral. And after integrating, this is going to be -4/7 of you to the negative 7/4 between Epsilon and 3. And now in that case if we apply our bones. Then this is going to be -4/7 of 3 to 74, OK. Plus 4/7. Of E to the 74. Now at this point, OK, as Epsilon approaches 0. From the right, OK. Then epsilon to the power of -7/4 will approach infinity. So that means this limit then is equal to infinity. And since it's equal to infinity, that means the limit diverges. Therefore, D is the correct answer. Thanks for watching everyone. I hope this video helped.

7–58. Improper integrals Evaluate the following integrals or state that they diverge.
44. ∫ (from 0 to ln 3) eʸ/(eʸ-1)⁷ᐟ³ dy

Key Concepts

Improper Integrals

Behavior of the Integrand Near Singularities

Substitution and Simplification Techniques

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