Question

Double integral question

Answer 1

\[\int\limits_{0}^{1}\int\limits_{x}^{2-x}\frac{ x }{ y } dydx\]

Answer 2

\[
\int_x^{2-x} \frac x y d y= x \ln (2-x)-x \ln (x)\\

\]

Answer 3

\[

\int_0^1( x \ln (2-x)-x \ln (x)) =\ln(4) -1
\]
The last one needs integration by parts

Answer 4

Thank you very much for replying with such detail !! I really appreciate it!

Answer 5

For example
\[
\int x \ln (x) \, dx=\frac{1}{2} x^2 \ln (x)-\frac{x^2}{4}\\
\int x \ln (2-x) \, dx=-\frac{x^2}{4}+\frac{1}{2} x^2 \ln (2-x)+x-2 \log
(2-x)

\]

Answer 6

YW

Answer 7

First note that \[\large \begin{aligned}\int_0^1\int_x^{2-x}\frac{x}{y}\,dy\,dx &= \int_0^1x\left.\left[\ln|y|\right]\right|_x^{2-x}\\ &= \int_0^1 x\ln\left|\frac{2-x}{x}\right|\,dx\\ &= \int_0^1 x\ln\left| \frac{2}{x}-1 \right|\,dx.\end{aligned}\]Now, we note that \(\large x\ln\left|\dfrac{2}{x}-1\right|\) is undefined at x=0; hence the resulting integral is improper. Furthemore, for \(0<x\leq 1\), we have \(\dfrac{2}{x}-1>0\), so we can drop absolute values. Thus, to integrate \(\large\displaystyle\int_0^1 x\ln\left(\dfrac{2}{x}-1\right)\,dx\), we proceed by integration by parts. Let \(\large\displaystyle u=\ln\left(\dfrac{2}{x}-1\right)\implies \,du=\frac{1}{\dfrac{2}{x}-1}\cdot-\dfrac{2}{x^2}\,dx = -\dfrac{2}{2x-x^2}\,dx\) and \(\large \,dv=x\,dx \implies v=\frac{1}{2}x^2\). Therefore,\[\begin{aligned}\int_0^1x\ln\left(\frac{2}{x}-1\right)\,dx &= \lim_{a\to 0^+}\left[\left.\left(\frac{1}{2}x^2\ln\left(\frac{2}{x}-1\right)\right)\right|_a^1 + \frac{1}{2}\int_a^1 \frac{2x^2}{x(2-x)}\,dx\right]\\ &= \lim_{a\to 0^+}\left[\left.\left(\frac{1}{2}x^2\ln\left(\frac{2}{x}-1\right)\right)\right|_a^1 + \int_a^1 \frac{x}{2-x}\,dx\right]\\ &= \lim_{a\to 0^+}\left[\left.\left(\frac{1}{2}x^2\ln\left(\frac{2}{x}-1\right)\right)\right|_a^1 - \int_a^1 1-\frac{2}{2-x}\,dx\right]\\ &= \lim_{a\to 0^+}\left.\left(\frac{1}{2}x^2\ln\left(\frac{2}{x}-1\right) - x - 2\ln|2-x|\right)\right|_a^1\\ &= -1 -\lim_{a\to 0^+}\left(\underbrace{\frac{a^2}{2}\ln\left(\frac{2}{a}-1\right)}_{=0\text{ by L'Hopital's Rule}} - a - 2\ln(2-a) \right)\\ &= -1 +2\ln 2 \\ &= \ln 4 - 1\end{aligned} \](I clearly got ninja'd by @eliassaab, but I didn't want this work to go to waste [especially after chrome crashed on my end and I had to restart my post from scratch]... XD)

I hope this makes sense! :-)

Answer 8

Wow! That is an incredibly detailed post! I really thank you for taking the time to write that for me. It definitely made sense so thanks a lot!!!!