Question Number 70872 by Abdo msup. last updated on 09/Oct/19
![calculate ∫∫_([1,3]^2 ) e^(−x−y) ln(2x+y)dxdy](Q70872.png)
$${calculate}\:\int\int_{\left[\mathrm{1},\mathrm{3}\right]^{\mathrm{2}} } \:\:\:{e}^{−{x}−{y}} \:{ln}\left(\mathrm{2}{x}+{y}\right){dxdy} \\ $$
Commented by Abdo msup. last updated on 11/Oct/19
![let I=∫∫_([1,3]^2 ) e^(−(x+y)) ln(2x+y)dxdy let consider the diffeomorphism (u,v)→(x,y)=(ϕ_1 (u,v),ϕ_2 (u,v)) /x+y=u and 2x+y=v ⇒x=−u+v and y=2u−v ⇒ ϕ_(1(u,v)) =−u+v and ϕ_2 (u,v)=2u−v M_j (ϕ)= ((((∂ϕ_1 /∂u) (∂ϕ_1 /∂v))),(((∂ϕ_2 /∂u) (∂ϕ_2 /∂v))) ) = (((−1 1)),((2 −1)) ) and detM_j =−1 ⇒ we have 1≤x≤3 and 1≤y≤3 ⇒2≤x+y≤6 ⇒ 2≤u≤6 2≤2x≤6 ⇒3≤2x+y≤9 ⇒3≤v≤9 ⇒ I=∫∫_(2≤u≤6 and 3≤v≤9) e^(−u) lnv du dv =∫_2 ^6 e^(−u) du .∫_3 ^9 lnv dv =[−e^(−u) ]_2 ^6 .[vlnv−v]_3 ^9 =( e^(−2) −e^(−6) )(9ln(9)−9−3ln(3)+3) =(e^(−2) −e^(−6) )(15ln(3)−6)](Q71026.png)
$${let}\:{I}=\int\int_{\left[\mathrm{1},\mathrm{3}\right]^{\mathrm{2}} } \:\:{e}^{−\left({x}+{y}\right)} {ln}\left(\mathrm{2}{x}+{y}\right){dxdy}\:{let}\: \\ $$$${consider}\:{the}\:{diffeomorphism} \\ $$$$\left({u},{v}\right)\rightarrow\left({x},{y}\right)=\left(\varphi_{\mathrm{1}} \left({u},{v}\right),\varphi_{\mathrm{2}} \left({u},{v}\right)\right)\:/{x}+{y}={u}\:{and} \\ $$$$\mathrm{2}{x}+{y}={v}\:\Rightarrow{x}=−{u}+{v}\:\:{and}\:{y}=\mathrm{2}{u}−{v}\:\Rightarrow \\ $$$$\varphi_{\mathrm{1}\left({u},{v}\right)} =−{u}+{v}\:{and}\:\varphi_{\mathrm{2}} \left({u},{v}\right)=\mathrm{2}{u}−{v} \\ $$$${M}_{{j}} \left(\varphi\right)=\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial{v}}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial{v}}}\end{pmatrix} \\ $$$$=\:\begin{pmatrix}{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}\:\:\:\:{and}\:{detM}_{{j}} =−\mathrm{1}\:\Rightarrow \\ $$$${we}\:{have}\:\mathrm{1}\leqslant{x}\leqslant\mathrm{3}\:{and}\:\mathrm{1}\leqslant{y}\leqslant\mathrm{3}\:\Rightarrow\mathrm{2}\leqslant{x}+{y}\leqslant\mathrm{6}\:\Rightarrow \\ $$$$\mathrm{2}\leqslant{u}\leqslant\mathrm{6} \\ $$$$\mathrm{2}\leqslant\mathrm{2}{x}\leqslant\mathrm{6}\:\Rightarrow\mathrm{3}\leqslant\mathrm{2}{x}+{y}\leqslant\mathrm{9}\:\Rightarrow\mathrm{3}\leqslant{v}\leqslant\mathrm{9}\:\Rightarrow \\ $$$${I}=\int\int_{\mathrm{2}\leqslant{u}\leqslant\mathrm{6}\:\:{and}\:\mathrm{3}\leqslant{v}\leqslant\mathrm{9}} \:\:\:\:{e}^{−{u}} {lnv}\:{du}\:{dv} \\ $$$$=\int_{\mathrm{2}} ^{\mathrm{6}} \:{e}^{−{u}} {du}\:.\int_{\mathrm{3}} ^{\mathrm{9}} {lnv}\:{dv} \\ $$$$=\left[−{e}^{−{u}} \right]_{\mathrm{2}} ^{\mathrm{6}} .\left[{vlnv}−{v}\right]_{\mathrm{3}} ^{\mathrm{9}} \\ $$$$=\left(\:{e}^{−\mathrm{2}} −{e}^{−\mathrm{6}} \right)\left(\mathrm{9}{ln}\left(\mathrm{9}\right)−\mathrm{9}−\mathrm{3}{ln}\left(\mathrm{3}\right)+\mathrm{3}\right) \\ $$$$=\left({e}^{−\mathrm{2}} −{e}^{−\mathrm{6}} \right)\left(\mathrm{15}{ln}\left(\mathrm{3}\right)−\mathrm{6}\right) \\ $$$$ \\ $$