calculate ∫∫_(0≤x≤1 and 1≤y≤2) e^(x/y) dxdy


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Question Number 46847 by maxmathsup by imad last updated on 01/Nov/18

$c a l c u l a t e \int \int_{0 ⩽ x ⩽ 1 a n d 1 ⩽ y ⩽ 2} e^{\frac{x}{y}} d x d y$
Commented by maxmathsup by imad last updated on 04/Nov/18
$I =∫_1 ^2 (∫_0 ^1 e^(x/y) dx)dy but ∫_0 ^1 e^(x/y) dx =[y e^(x/y) ]_(x=0) ^(x =1) = y(e^(1/y) −1) ⇒ I =∫_1 ^2 (y e^(1/y) )dy −∫_1 ^2 ydy but ∫_1 ^2 ydy =[(y^2 /2)]_1 ^2 =2−(1/2) =(3/2) and ∫_1 ^2 y e^(1/y) dy =_((1/y)=t) ∫_(1/2) ^1 (1/t) e^t (dt/t^2 ) =∫_(1/2) ^1 (e^t /t^3 )dt =∫_(1/2) ^1 t^(−3) e^t dt =[−(1/2) t^(−2) e^t ]_(1/2) ^1 +∫_(1/2) ^1 (1/2) t^(−2) e^t dt =(1/2){ 4 e^(1/2) −e} +(1/2){ [−(1/t) e^t ]_(1/2) ^1 +∫_(1/2) ^1 (1/t) e^t dt} 2(√e)−(e/2) +(1/2){2(√e)−e + ∫_(1/2) ^1 (e^t /t) dt} =3(√e)−e +(1/2) ∫_(1/2) ^1 (e^t /t) dt and ∫_(1/2) ^1 (e^t /t) dt =∫_(1/2) ^1 (1/t)(Σ_(n=0) ^∞ (t^n /(n!)))dt =∫_(1/2) ^1 (1/t)(1+Σ_(n=1) ^∞ (t^n /(n!)))dt =ln(2) +Σ_(n=1) ^∞ (1/(n!)) ∫_(1/2) ^1 t^(n−1) dt =ln(2)+Σ_(n=1) ^∞ (1/(n.(n!)))(1−(1/2^n )) ...be continued...$
$I = \int_{1}^{2} (\int_{0}^{1} e^{\frac{x}{y}} d x) d y b u t \int_{0}^{1} e^{\frac{x}{y}} d x = {[y e^{\frac{x}{y}}]}_{x = 0}^{x = 1} = y (e^{\frac{1}{y}} - 1) \Rightarrow$ $I = \int_{1}^{2} (y e^{\frac{1}{y}}) d y - \int_{1}^{2} y d y b u t \int_{1}^{2} y d y = {[\frac{y^{2}}{2}]}_{1}^{2} = 2 - \frac{1}{2} = \frac{3}{2} a n d$ $\int_{1}^{2} y e^{\frac{1}{y}} d y =_{\frac{1}{y} = t} \int_{\frac{1}{2}}^{1} \frac{1}{t} e^{t} \frac{d t}{t^{2}} = \int_{\frac{1}{2}}^{1} \frac{e^{t}}{t^{3}} d t = \int_{\frac{1}{2}}^{1} t^{- 3} e^{t} d t$ $= {[- \frac{1}{2} t^{- 2} e^{t}]}_{\frac{1}{2}}^{1} + \int_{\frac{1}{2}}^{1} \frac{1}{2} t^{- 2} e^{t} d t$ $= \frac{1}{2} {4 e^{\frac{1}{2}} - e} + \frac{1}{2} {{[- \frac{1}{t} e^{t}]}_{\frac{1}{2}}^{1} + \int_{\frac{1}{2}}^{1} \frac{1}{t} e^{t} d t}$ $2 \sqrt{e} - \frac{e}{2} + \frac{1}{2} {2 \sqrt{e} - e + \int_{\frac{1}{2}}^{1} \frac{e^{t}}{t} d t}$ $= 3 \sqrt{e} - e + \frac{1}{2} \int_{\frac{1}{2}}^{1} \frac{e^{t}}{t} d t a n d \int_{\frac{1}{2}}^{1} \frac{e^{t}}{t} d t = \int_{\frac{1}{2}}^{1} \frac{1}{t} (\sum_{n = 0}^{\infty} \frac{t^{n}}{n!}) d t$ $= \int_{\frac{1}{2}}^{1} \frac{1}{t} (1 + \sum_{n = 1}^{\infty} \frac{t^{n}}{n!}) d t = l n (2) + \sum_{n = 1}^{\infty} \frac{1}{n!} \int_{\frac{1}{2}}^{1} t^{n - 1} d t$ $= l n (2) + \sum_{n = 1}^{\infty} \frac{1}{n . (n!)} (1 - \frac{1}{2^{n}}) . . . b e c o n t i n u e d . . .$
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