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Problem: Find $$\int_1^e\int_0^{\ln x}xy\,dy\,dx$$ with the given order of integration and with the order of integration reversed.

My Attempt: When I solve this problem with the given order of integration, I get $$\frac{e^2-1}{8}$$ and when I solve this problem with the order of integration reversed I get the integral $$\int_0^1\int_0^{e-e^{y}}xy\,dx\,dy=\frac{1}{2}\left(\frac{3e^2}{4}+\frac{1}{4}-2e\right).$$ The answers are different and so I must be wrong somewhere. Please point my mistake or provide me with a meaningful hint.

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    $\begingroup$ I get limits of $e^y$ and $e$ on the inner integral. $\endgroup$ Commented May 4, 2017 at 17:23
  • $\begingroup$ @LordSharktheUnknown In that case, I get the result to be $\frac{3e^2+1}{8}.$ $\endgroup$
    – Student
    Commented May 4, 2017 at 17:34
  • $\begingroup$ @LordSharktheUnknown. Shrey, can you see why? $\endgroup$ Commented May 4, 2017 at 17:34
  • $\begingroup$ No, it gives the same. $\endgroup$ Commented May 4, 2017 at 17:39
  • $\begingroup$ Yes, it gives the same. Thank you! $\endgroup$
    – Student
    Commented May 4, 2017 at 17:41

2 Answers 2

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$\int_1^e\int_0^{\ln x}xy\,dy\,dx\\ \int_1^e \frac 12 xy^2 |_0^{\ln x}dx\\ \int_1^e \frac 12 x(\ln x)^2dx\\ \frac 14 x^2\ln^2 x - \frac14 x^2\ln x + \frac 18 x^2 |_1^e\\ \frac 14 e^2 - \frac14 e^2 + \frac 18 e^2 - \frac 18\\ \frac 18 (e^2 - 1) $

that part looks right

now flipping the order of integration

the region is bounded by:

$y = \ln x\\ y = 0\\ x = e$

$\int_0^1\int_{e^y}^e xy \,dx\, dy\\ \int_0^1\frac 12 y (e^2 - e^{2y})\, dy\\ \frac 14 e^2 y^2 - \frac 18 e^{2y}(2y-1)|_0^1\\ \frac 14e^2 - \frac 18 e^2 - \frac 18\\ \frac 18 (e^2 - 1) $

Where do you think y0u went wrong?

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The original integral is $\int_0^e\int_0^{ln(y)} xy dy dx$. The limits of integration are: x goes from 0 to e and, for every x, y goes from 0 to ln(x). Drawing a graph of y= ln(x), we see that it goes below the x-axis for x between 0 and 1. To reverse the order of integration, we need to break the integral into 2 parts, y< 0 and y> 0. For y< 0, x goes form 0 to $e^y$. For y> 0, x goes from e^y to e. The integral is $\int_{-\infty}^0 \int_{e^y}^1 xy dxdy+ \int_0^1\int_{e^y}^e xy dxdy$

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  • $\begingroup$ I don't know what means "original" in this context. It cannot mean "the one the OP wrote" because he didn't wrote that. I'm confused. $\endgroup$ Commented May 4, 2017 at 19:24
  • $\begingroup$ You are right- the integral with respect to x is from 1 to e, not 0 to e. Then, of course, what I said about two different integrals does not apply. In the integral the OP wrote, x goes form 1 to e and, for every e, y goes from 0 to ln(x). To reverse the order of integration, y must go form 0 to ln(e)= 1 and, for each y, x goes $e^y$ to 1. The integral is $\int_0^1\int_{e^y}^e xy dxdy$. $\endgroup$
    – user247327
    Commented May 4, 2017 at 19:49
  • $\begingroup$ What confused me was the answer accepted :) $\endgroup$ Commented May 4, 2017 at 19:55

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