Question Number 192743 by mathocean1 last updated on 25/May/23
$${Calculate}: \\ $$$${I}=\int_{\mathrm{0}} ^{\:\mathrm{4}} \left(\int_{{x}−\mathrm{4}} ^{\:{x}−\mathrm{2}} {e}^{\frac{{x}+{y}}{{x}−{y}}} {dy}\right){dx} \\ $$
Answered by aleks041103 last updated on 27/May/23
![{ ((X=x+y)),((Y=x−y)) :} ⇒ { ((x=(1/2)(X+Y))),((y=(1/2)(X−Y))) :} J= (((∂x/∂X),(∂x/∂Y)),((∂y/∂X),(∂y/∂Y)) ) = (((1/2),(1/2)),((1/2),((−1)/2)) ) ⇒∣J∣=−(1/2) ⇒I=∫_(G′) e^(X/Y) ∣∣J∣∣dXdY=(1/2)∫_(G′) e^(X/Y) dXdY G: { ((0≤x≤4)),((x−4≤y≤x−2)) :} ⇒G′: { ((0≤(1/2)(X+Y)≤4)),(((1/2)(X+Y)−4≤(1/2)(X−Y)≤(1/2)(X+Y)−2)) :} ⇒G′: { ((−X≤Y≤8−X)),((Y−8≤−Y≤Y−4)) :}⇒G′: { ((2≤Y≤4)),((−Y≤X≤8−Y)) :} ⇒I=(1/2)∫_2 ^4 (∫_(−y) ^( 8−y) e^(x/y) dx)dy ∫_(−y) ^( 8−y) e^(x/y) dx=[ye^(x/y) ]_(−y) ^(8−y) =y(e^(8/y−1) −e^(−1) )= =(y/e)(e^(8/y) −1) I=(1/(2e))∫_2 ^( 4) y(e^(8/y) −1)dy= =(1/(2e))(∫_2 ^( 4) ye^(8/y) dy−[(y^2 /2)]_2 ^4 )= =(1/(2e))(∫_2 ^( 4) ye^(8/y) dy−6) z=8/y⇒y=8/z⇒−dy=(8/z^2 )dz ∫_2 ^( 4) ye^(8/y) dy=−64∫_4 ^( 2) z^(−3) e^z dz=64∫_2 ^( 4) z^(−3) e^z dz ⇒I=(1/e)(32∫_2 ^( 4) z^(−3) e^z dz−3) ∫t^(−3) e^t dt= =(e^t /t^2 )−∫td(t^(−3) e^t )= =(e^t /t^2 )−∫(−3t^(−3) e^t +t^(−2) e^t )dt= =t^(−2) e^t +3∫t^(−3) e^t dt−∫t^(−2) e^t dt ⇒∫t^(−3) e^t dt=(1/2)(−t^(−2) e^t +∫t^(−2) e^t dt) ∫t^(−2) e^t dt=t^(−1) e^t −∫td(t^(−2) e^t )= =t^(−1) e^t −∫(−2t^(−2) e^t +t^(−1) e^t )dt= =t^(−1) e^t +2∫t^(−2) e^t dt−∫t^(−1) e^t dt ⇒∫t^(−2) e^t dt=−((e^t /t)−∫ ((e^t dt)/t))= =−(e^t /t)+Ei(t) ⇒∫t^(−3) e^t dt=(1/2)(−t^(−2) e^t +∫t^(−2) e^t dt)= =(1/2)(−(e^t /t)−(e^t /t^2 )+Ei(t))= =(1/2)(−(((t+1)e^t )/t^2 )+Ei(t)) ⇒∫_2 ^( 4) x^(−3) e^x dx=(1/2)[−(((x+1)e^x )/x^2 )+Ei(x)]_2 ^4 = =(1/2)(((3e^2 )/4)−((5e^4 )/(16))+Ei(4)−Ei(2)) ⇒I=(1/e)(16(((3e^2 )/4)−((5e^4 )/(16))+Ei(4)−Ei(2))−3)= =(1/e)(12e^2 −5e^4 −3+16Ei(4)−16Ei(2)) I=∫_0 ^( 4) ∫_(x−4) ^( x−2) e^((x+y)/(x−y)) dydx=(1/e)(12e^2 −5e^4 −3+16Ei(4)−16Ei(2)) I≈17.4758](https://www.tinkutara.com/question/Q192790.png)
$$\begin{cases}{{X}={x}+{y}}\\{{Y}={x}−{y}}\end{cases}\:\Rightarrow\begin{cases}{{x}=\frac{\mathrm{1}}{\mathrm{2}}\left({X}+{Y}\right)}\\{{y}=\frac{\mathrm{1}}{\mathrm{2}}\left({X}−{Y}\right)}\end{cases} \\ $$$${J}=\begin{pmatrix}{\frac{\partial{x}}{\partial{X}}}&{\frac{\partial{x}}{\partial{Y}}}\\{\frac{\partial{y}}{\partial{X}}}&{\frac{\partial{y}}{\partial{Y}}}\end{pmatrix}\:=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}&{\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}}&{\frac{−\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\Rightarrow\mid{J}\mid=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{I}=\int_{{G}'} {e}^{{X}/{Y}} \mid\mid{J}\mid\mid{dXdY}=\frac{\mathrm{1}}{\mathrm{2}}\int_{{G}'} {e}^{{X}/{Y}} {dXdY} \\ $$$${G}:\begin{cases}{\mathrm{0}\leqslant{x}\leqslant\mathrm{4}}\\{{x}−\mathrm{4}\leqslant{y}\leqslant{x}−\mathrm{2}}\end{cases} \\ $$$$\Rightarrow{G}':\begin{cases}{\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({X}+{Y}\right)\leqslant\mathrm{4}}\\{\frac{\mathrm{1}}{\mathrm{2}}\left({X}+{Y}\right)−\mathrm{4}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({X}−{Y}\right)\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({X}+{Y}\right)−\mathrm{2}}\end{cases} \\ $$$$\Rightarrow{G}':\begin{cases}{−{X}\leqslant{Y}\leqslant\mathrm{8}−{X}}\\{{Y}−\mathrm{8}\leqslant−{Y}\leqslant{Y}−\mathrm{4}}\end{cases}\Rightarrow{G}':\begin{cases}{\mathrm{2}\leqslant{Y}\leqslant\mathrm{4}}\\{−{Y}\leqslant{X}\leqslant\mathrm{8}−{Y}}\end{cases} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{2}} ^{\mathrm{4}} \left(\int_{−{y}} ^{\:\mathrm{8}−{y}} {e}^{{x}/{y}} {dx}\right){dy} \\ $$$$\int_{−{y}} ^{\:\mathrm{8}−{y}} {e}^{{x}/{y}} {dx}=\left[{ye}^{{x}/{y}} \right]_{−{y}} ^{\mathrm{8}−{y}} ={y}\left({e}^{\mathrm{8}/{y}−\mathrm{1}} −{e}^{−\mathrm{1}} \right)= \\ $$$$=\frac{{y}}{{e}}\left({e}^{\mathrm{8}/{y}} −\mathrm{1}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}{e}}\int_{\mathrm{2}} ^{\:\mathrm{4}} {y}\left({e}^{\mathrm{8}/{y}} −\mathrm{1}\right){dy}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{e}}\left(\int_{\mathrm{2}} ^{\:\mathrm{4}} {ye}^{\mathrm{8}/{y}} {dy}−\left[\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{2}} ^{\mathrm{4}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{e}}\left(\int_{\mathrm{2}} ^{\:\mathrm{4}} {ye}^{\mathrm{8}/{y}} {dy}−\mathrm{6}\right) \\ $$$${z}=\mathrm{8}/{y}\Rightarrow{y}=\mathrm{8}/{z}\Rightarrow−{dy}=\frac{\mathrm{8}}{{z}^{\mathrm{2}} }{dz} \\ $$$$\int_{\mathrm{2}} ^{\:\mathrm{4}} {ye}^{\mathrm{8}/{y}} {dy}=−\mathrm{64}\int_{\mathrm{4}} ^{\:\mathrm{2}} {z}^{−\mathrm{3}} {e}^{{z}} {dz}=\mathrm{64}\int_{\mathrm{2}} ^{\:\mathrm{4}} {z}^{−\mathrm{3}} {e}^{{z}} {dz} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{{e}}\left(\mathrm{32}\int_{\mathrm{2}} ^{\:\mathrm{4}} {z}^{−\mathrm{3}} {e}^{{z}} {dz}−\mathrm{3}\right) \\ $$$$\int{t}^{−\mathrm{3}} {e}^{{t}} {dt}= \\ $$$$=\frac{{e}^{{t}} }{{t}^{\mathrm{2}} }−\int{td}\left({t}^{−\mathrm{3}} {e}^{{t}} \right)= \\ $$$$=\frac{{e}^{{t}} }{{t}^{\mathrm{2}} }−\int\left(−\mathrm{3}{t}^{−\mathrm{3}} {e}^{{t}} +{t}^{−\mathrm{2}} {e}^{{t}} \right){dt}= \\ $$$$={t}^{−\mathrm{2}} {e}^{{t}} +\mathrm{3}\int{t}^{−\mathrm{3}} {e}^{{t}} {dt}−\int{t}^{−\mathrm{2}} {e}^{{t}} {dt} \\ $$$$\Rightarrow\int{t}^{−\mathrm{3}} {e}^{{t}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(−{t}^{−\mathrm{2}} {e}^{{t}} +\int{t}^{−\mathrm{2}} {e}^{{t}} {dt}\right) \\ $$$$\int{t}^{−\mathrm{2}} {e}^{{t}} {dt}={t}^{−\mathrm{1}} {e}^{{t}} −\int{td}\left({t}^{−\mathrm{2}} {e}^{{t}} \right)= \\ $$$$={t}^{−\mathrm{1}} {e}^{{t}} −\int\left(−\mathrm{2}{t}^{−\mathrm{2}} {e}^{{t}} +{t}^{−\mathrm{1}} {e}^{{t}} \right){dt}= \\ $$$$={t}^{−\mathrm{1}} {e}^{{t}} +\mathrm{2}\int{t}^{−\mathrm{2}} {e}^{{t}} {dt}−\int{t}^{−\mathrm{1}} {e}^{{t}} {dt} \\ $$$$\Rightarrow\int{t}^{−\mathrm{2}} {e}^{{t}} {dt}=−\left(\frac{{e}^{{t}} }{{t}}−\int\:\frac{{e}^{{t}} {dt}}{{t}}\right)= \\ $$$$=−\frac{{e}^{{t}} }{{t}}+{Ei}\left({t}\right) \\ $$$$\Rightarrow\int{t}^{−\mathrm{3}} {e}^{{t}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(−{t}^{−\mathrm{2}} {e}^{{t}} +\int{t}^{−\mathrm{2}} {e}^{{t}} {dt}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{{e}^{{t}} }{{t}}−\frac{{e}^{{t}} }{{t}^{\mathrm{2}} }+{Ei}\left({t}\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\left({t}+\mathrm{1}\right){e}^{{t}} }{{t}^{\mathrm{2}} }+{Ei}\left({t}\right)\right) \\ $$$$\Rightarrow\int_{\mathrm{2}} ^{\:\mathrm{4}} {x}^{−\mathrm{3}} {e}^{{x}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[−\frac{\left({x}+\mathrm{1}\right){e}^{{x}} }{{x}^{\mathrm{2}} }+{Ei}\left({x}\right)\right]_{\mathrm{2}} ^{\mathrm{4}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}{e}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{5}{e}^{\mathrm{4}} }{\mathrm{16}}+{Ei}\left(\mathrm{4}\right)−{Ei}\left(\mathrm{2}\right)\right) \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{{e}}\left(\mathrm{16}\left(\frac{\mathrm{3}{e}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{5}{e}^{\mathrm{4}} }{\mathrm{16}}+{Ei}\left(\mathrm{4}\right)−{Ei}\left(\mathrm{2}\right)\right)−\mathrm{3}\right)= \\ $$$$=\frac{\mathrm{1}}{{e}}\left(\mathrm{12}{e}^{\mathrm{2}} −\mathrm{5}{e}^{\mathrm{4}} −\mathrm{3}+\mathrm{16}{Ei}\left(\mathrm{4}\right)−\mathrm{16}{Ei}\left(\mathrm{2}\right)\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\:\mathrm{4}} \int_{{x}−\mathrm{4}} ^{\:{x}−\mathrm{2}} {e}^{\frac{{x}+{y}}{{x}−{y}}} {dydx}=\frac{\mathrm{1}}{{e}}\left(\mathrm{12}{e}^{\mathrm{2}} −\mathrm{5}{e}^{\mathrm{4}} −\mathrm{3}+\mathrm{16}{Ei}\left(\mathrm{4}\right)−\mathrm{16}{Ei}\left(\mathrm{2}\right)\right) \\ $$$${I}\approx\mathrm{17}.\mathrm{4758} \\ $$
Commented by mathocean1 last updated on 27/May/23
$${Thank}\:{you}\:{very}\:{much} \\ $$