calculate-0-x-1-and-1-y-2-e-x-y-dxdy-

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 = \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}}) \dots b e c o n t i n u e d \dots