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Math Help - Double integral of e^(x/y)

  1. #1
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    Double integral of e^(x/y)

    Can someone help me get at least a start with the double integral? I am totally stumped.

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

    Thanks in advance.
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  2. #2
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    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 \begin{align*} 0 \leq x \leq 1 \end{align*} and \displaystyle \begin{align*} x \leq y \leq 1 \end{align*}. Therefore \displaystyle \begin{align*} 0 \leq x \leq y \leq 1 \end{align*}. So that means we can change the bounds to \displaystyle \begin{align*} 0 \leq y \leq 1 \end{align*} and \displaystyle \begin{align*} 0 \leq x \leq y \end{align*}, giving

    \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
    Thanks from topsquark
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  3. #3
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    Re: Double integral of e^(x/y)

    Thanks for the help. Would it be possible to check my steps?

    \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}
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  4. #4
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    Re: Double integral of e^(x/y)

    It looks fine, well done
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