# Double integral of e^(x/y)

• Apr 30th 2013, 02:24 AM
tammyl
Double integral of e^(x/y)
Can someone help me get at least a start with the double integral? I am totally stumped.

$\displaystyle \int_0^1 \int_x^1 e^\frac{x}{y} dy dx$

• Apr 30th 2013, 02:39 AM
Prove It
Re: Double integral of e^(x/y)
It would be a good idea to reverse the order of integration. Notice from your bounds that \displaystyle \displaystyle \begin{align*} 0 \leq x \leq 1 \end{align*} and \displaystyle \displaystyle \begin{align*} x \leq y \leq 1 \end{align*}. Therefore \displaystyle \displaystyle \begin{align*} 0 \leq x \leq y \leq 1 \end{align*}. So that means we can change the bounds to \displaystyle \displaystyle \begin{align*} 0 \leq y \leq 1 \end{align*} and \displaystyle \displaystyle \begin{align*} 0 \leq x \leq y \end{align*}, giving

\displaystyle \displaystyle \begin{align*} \int_0^1{\int_x^1{e^{\frac{x}{y}}\,dy}\,dx} = \int_0^1{\int_0^y{e^{\frac{x}{y}}\,dx}\,dy} \end{align*}

See how you go from here :)
• Apr 30th 2013, 03:23 AM
tammyl
Re: Double integral of e^(x/y)
Thanks for the help. Would it be possible to check my steps?

$\displaystyle \int_0^1 \int_x^1 e^\frac{x}{y} dy dx = \int_0^1 \int_0^y e^\frac{x}{y} dx dy \\ = \int_0^1 [y e^\frac{x}{y}]_0^y dy \\ = \int_0^1 [y e^\frac{y}{y} - y e^\frac{0}{y}] dy \\ = \int_0^1 [y e - y] dy \\ = \int_0^1 [y (e - 1)] dy \\ = (e-1) \int_0^1 y dy \\ = (e-1) [\frac{y^2}{2}]_0^1 \\ = (e-1) (\frac{1^2}{2} - \frac{0^2}{2}) \\ = \frac {e-1}{2}$
• Apr 30th 2013, 03:24 AM
Prove It
Re: Double integral of e^(x/y)
It looks fine, well done :)