Calculate: I=∫_0 ^( 4) (∫_(x−4) ^( x−2) e^((x+y)/(x−y)) dy)dx


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 192743 by mathocean1 last updated on 25/May/23

$C a l c u l a t e :$ $I = \int_{0}^{4} (\int_{x - 4}^{x - 2} e^{\frac{x + y}{x - y}} d y) d x$
	Answered by aleks041103 last updated on 27/May/23
	${ ((X=x+y)),((Y=x−y)) :} ⇒ { ((x=(1/2)(X+Y))),((y=(1/2)(X−Y))) :} J= (((∂x/∂X),(∂x/∂Y)),((∂y/∂X),(∂y/∂Y)) ) = (((1/2),(1/2)),((1/2),((−1)/2)) ) ⇒∣J∣=−(1/2) ⇒I=∫_(G′) e^(X/Y) ∣∣J∣∣dXdY=(1/2)∫_(G′) e^(X/Y) dXdY G: { ((0≤x≤4)),((x−4≤y≤x−2)) :} ⇒G′: { ((0≤(1/2)(X+Y)≤4)),(((1/2)(X+Y)−4≤(1/2)(X−Y)≤(1/2)(X+Y)−2)) :} ⇒G′: { ((−X≤Y≤8−X)),((Y−8≤−Y≤Y−4)) :}⇒G′: { ((2≤Y≤4)),((−Y≤X≤8−Y)) :} ⇒I=(1/2)∫_2 ^4 (∫_(−y) ^( 8−y) e^(x/y) dx)dy ∫_(−y) ^( 8−y) e^(x/y) dx=[ye^(x/y) ]_(−y) ^(8−y) =y(e^(8/y−1) −e^(−1) )= =(y/e)(e^(8/y) −1) I=(1/(2e))∫_2 ^( 4) y(e^(8/y) −1)dy= =(1/(2e))(∫_2 ^( 4) ye^(8/y) dy−[(y^2 /2)]_2 ^4 )= =(1/(2e))(∫_2 ^( 4) ye^(8/y) dy−6) z=8/y⇒y=8/z⇒−dy=(8/z^2 )dz ∫_2 ^( 4) ye^(8/y) dy=−64∫_4 ^( 2) z^(−3) e^z dz=64∫_2 ^( 4) z^(−3) e^z dz ⇒I=(1/e)(32∫_2 ^( 4) z^(−3) e^z dz−3) ∫t^(−3) e^t dt= =(e^t /t^2 )−∫td(t^(−3) e^t )= =(e^t /t^2 )−∫(−3t^(−3) e^t +t^(−2) e^t )dt= =t^(−2) e^t +3∫t^(−3) e^t dt−∫t^(−2) e^t dt ⇒∫t^(−3) e^t dt=(1/2)(−t^(−2) e^t +∫t^(−2) e^t dt) ∫t^(−2) e^t dt=t^(−1) e^t −∫td(t^(−2) e^t )= =t^(−1) e^t −∫(−2t^(−2) e^t +t^(−1) e^t )dt= =t^(−1) e^t +2∫t^(−2) e^t dt−∫t^(−1) e^t dt ⇒∫t^(−2) e^t dt=−((e^t /t)−∫ ((e^t dt)/t))= =−(e^t /t)+Ei(t) ⇒∫t^(−3) e^t dt=(1/2)(−t^(−2) e^t +∫t^(−2) e^t dt)= =(1/2)(−(e^t /t)−(e^t /t^2 )+Ei(t))= =(1/2)(−(((t+1)e^t )/t^2 )+Ei(t)) ⇒∫_2 ^( 4) x^(−3) e^x dx=(1/2)[−(((x+1)e^x )/x^2 )+Ei(x)]_2 ^4 = =(1/2)(((3e^2 )/4)−((5e^4 )/(16))+Ei(4)−Ei(2)) ⇒I=(1/e)(16(((3e^2 )/4)−((5e^4 )/(16))+Ei(4)−Ei(2))−3)= =(1/e)(12e^2 −5e^4 −3+16Ei(4)−16Ei(2)) I=∫_0 ^( 4) ∫_(x−4) ^( x−2) e^((x+y)/(x−y)) dydx=(1/e)(12e^2 −5e^4 −3+16Ei(4)−16Ei(2)) I≈17.4758$
	${\begin{cases} X = x + y \\ Y = x - y \end{cases} \Rightarrow {\begin{cases} x = \frac{1}{2} (X + Y) \\ y = \frac{1}{2} (X - Y) \end{cases}$ $J = (\begin{matrix} \frac{\partial x}{\partial X} & \frac{\partial x}{\partial Y} \\ \frac{\partial y}{\partial X} & \frac{\partial y}{\partial Y} \end{matrix}) = (\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{- 1}{2} \end{matrix})$ $\Rightarrow∣ J ∣= - \frac{1}{2}$ $\Rightarrow I = \int_{G^{'}} e^{X / Y} ∣∣ J ∣∣ d X d Y = \frac{1}{2} \int_{G^{'}} e^{X / Y} d X d Y$ $G : {\begin{cases} 0 ⩽ x ⩽ 4 \\ x - 4 ⩽ y ⩽ x - 2 \end{cases}$ $\Rightarrow G^{'} : {\begin{cases} 0 ⩽ \frac{1}{2} (X + Y) ⩽ 4 \\ \frac{1}{2} (X + Y) - 4 ⩽ \frac{1}{2} (X - Y) ⩽ \frac{1}{2} (X + Y) - 2 \end{cases}$ $\Rightarrow G^{'} : {\begin{cases} - X ⩽ Y ⩽ 8 - X \\ Y - 8 ⩽ - Y ⩽ Y - 4 \end{cases} \Rightarrow G^{'} : {\begin{cases} 2 ⩽ Y ⩽ 4 \\ - Y ⩽ X ⩽ 8 - Y \end{cases}$ $\Rightarrow I = \frac{1}{2} \int_{2}^{4} (\int_{- y}^{8 - y} e^{x / y} d x) d y$ $\int_{- y}^{8 - y} e^{x / y} d x = {[{y e}^{x / y}]}_{- y}^{8 - y} = y (e^{8 / y - 1} - e^{- 1}) =$ $= \frac{y}{e} (e^{8 / y} - 1)$ $I = \frac{1}{2 e} \int_{2}^{4} y (e^{8 / y} - 1) d y =$ $= \frac{1}{2 e} (\int_{2}^{4} {y e}^{8 / y} d y - {[\frac{y^{2}}{2}]}_{2}^{4}) =$ $= \frac{1}{2 e} (\int_{2}^{4} {y e}^{8 / y} d y - 6)$ $z = 8 / y \Rightarrow y = 8 / z \Rightarrow - d y = \frac{8}{z^{2}} d z$ $\int_{2}^{4} {y e}^{8 / y} d y = - 64 \int_{4}^{2} z^{- 3} e^{z} d z = 64 \int_{2}^{4} z^{- 3} e^{z} d z$ $\Rightarrow I = \frac{1}{e} (32 \int_{2}^{4} z^{- 3} e^{z} d z - 3)$ $\int t^{- 3} e^{t} d t =$ $= \frac{e^{t}}{t^{2}} - \int t d (t^{- 3} e^{t}) =$ $= \frac{e^{t}}{t^{2}} - \int (- 3 t^{- 3} e^{t} + t^{- 2} e^{t}) d t =$ $= t^{- 2} e^{t} + 3 \int t^{- 3} e^{t} d t - \int t^{- 2} e^{t} d t$ $\Rightarrow \int t^{- 3} e^{t} d t = \frac{1}{2} (- t^{- 2} e^{t} + \int t^{- 2} e^{t} d t)$ $\int t^{- 2} e^{t} d t = t^{- 1} e^{t} - \int t d (t^{- 2} e^{t}) =$ $= t^{- 1} e^{t} - \int (- 2 t^{- 2} e^{t} + t^{- 1} e^{t}) d t =$ $= t^{- 1} e^{t} + 2 \int t^{- 2} e^{t} d t - \int t^{- 1} e^{t} d t$ $\Rightarrow \int t^{- 2} e^{t} d t = - (\frac{e^{t}}{t} - \int \frac{e^{t} d t}{t}) =$ $= - \frac{e^{t}}{t} + E i (t)$ $\Rightarrow \int t^{- 3} e^{t} d t = \frac{1}{2} (- t^{- 2} e^{t} + \int t^{- 2} e^{t} d t) =$ $= \frac{1}{2} (- \frac{e^{t}}{t} - \frac{e^{t}}{t^{2}} + E i (t)) =$ $= \frac{1}{2} (- \frac{(t + 1) e^{t}}{t^{2}} + E i (t))$ $\Rightarrow \int_{2}^{4} x^{- 3} e^{x} d x = \frac{1}{2} {[- \frac{(x + 1) e^{x}}{x^{2}} + E i (x)]}_{2}^{4} =$ $= \frac{1}{2} (\frac{3 e^{2}}{4} - \frac{5 e^{4}}{16} + E i (4) - E i (2))$ $\Rightarrow I = \frac{1}{e} (16 (\frac{3 e^{2}}{4} - \frac{5 e^{4}}{16} + E i (4) - E i (2)) - 3) =$ $= \frac{1}{e} (12 e^{2} - 5 e^{4} - 3 + 16 E i (4) - 16 E i (2))$ $I = \int_{0}^{4} \int_{x - 4}^{x - 2} e^{\frac{x + y}{x - y}} d y d x = \frac{1}{e} (12 e^{2} - 5 e^{4} - 3 + 16 E i (4) - 16 E i (2))$ $I \approx 17.4758$
	Commented by mathocean1 last updated on 27/May/23

	$T h a n k y o u v e r y m u c h$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum