Calculate $\iint_D x\ln(xy) dx\,dy \text{ where } x = 1, x = e, y = \frac{2}{x}, y = \frac{1}{x}$
Ok this is a sample exercise from the book that I don't know how to solve.
Calculate $\iint_D x\ln(xy) dx\,dy \text{ where } 1 \le x \le e , \frac{2}{x} \le y \le \frac{1}{x}$
The answer is $(\ln4 -1)(e - 1)$ but I can't figure out why.
I thought to change order or integration but I'm not successing to do it.
Can you please help me solve it? thanks.
$\endgroup$ 63 Answers
$\begingroup$$$ \int_1^\text{e}\left( \int_\frac{2}{x}^\frac{1}{x}x \ln(xy)\;\mathrm{d}y\right)\;\mathrm{d}x $$ To make the computations more clear lets figure out the first integrand first. $$ \int_\frac{2}{x}^\frac{1}{x}x \ln(xy)\;\mathrm{d}y = \left[ \begin{array}{c c} u = \ln(xy)& v' = 1\\ u' = \frac{1}{xy}\cdot x & v= y \end{array} \right] = x\left(\left[ y\cdot\ln(xy) \right]_{2/x}^{1/x} -\int_\frac{2}{x}^\frac{1}{x}\frac{1}{y}\cdot y\; \mathrm{d}y \right) = x\left( \left(\frac{\ln 1}{x}\right)-\left(\frac{\ln 2}{x}\right)-\left[y\right]_{2/x}^{1/x} \right) = x\left(-\frac{1+\ln 2}{x}\right) = -1-\ln 2 $$ And then just put it into the first integrand and integrate once more. $$ \int_1^\text{e} -1 -\ln 2\; \mathrm{d}x = -(1+\ln 2)\left[x\right]_1^\text{e} = -(1+\ln 2)(e-1) $$ Well, it seems I've made some numeric mistakes, but hopefully the integrating process is clear.
$\endgroup$ $\begingroup$First of all, it's better if you integrate on $y$ first and on $x$ second. Now, you have the integral $$\int_1^e \left(\int_{1/x}^{2/x} x \ln(xy) dy\right)dx.$$
Carry the $x$ out into the second integral and calculate the integral $\int_{1/x}^{2/x}\ln(xy) dy$. You get some function of $x$ which you can then integrate.
$\endgroup$ $\begingroup$Make a change of variables $$\varphi\colon\;\;\pmatrix{x\\y}\mapsto \pmatrix{u\\v}$$ by $$\begin{cases} u=x,\\ v= xy. \end{cases}$$ Then $$\dfrac{\partial(u,\,v)}{\partial(x,\,y)}=\pmatrix{ \dfrac{\partial{u}}{\partial{x}}& \dfrac{\partial{u}}{\partial{y}}\\\dfrac{\partial{v}}{\partial{x}} & \dfrac{\partial{v}}{\partial{y}}}=\pmatrix{y&x\\1&0},$$ so $$\det{\dfrac{\partial(u,\,v)}{\partial(x,\,y)}}=x\ne{0}$$ in $D,$ therefore the Jacobian determinant $$\det{\dfrac{\partial(x,\,y)}{\partial(u,\,v)}}=\dfrac{1}{u}.$$ Additionally, $D'=\varphi(D)=\{(u,\,v)\colon\;\; 1\leqslant{u} \leqslant{e},\;\;{1\leqslant{v}\leqslant{2}}\}.$ Then $$\iint\limits_{D} {x\ln(xy)} dx\,dy=\iint\limits_{D'} {u\ln{(v)}} \dfrac{1}{u}\ du\,dv= \int\limits_{1}^{e}{\left(\int\limits_{1}^{2}{\ln{(v)}\ dv}\right)\ du}$$
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