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Math Help - Double integral problem, help please

  1. #1
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    Double integral problem, help please

    Double integral - e^(x/y) dA D = 1<y<2, y<x<y^3 (all the inequalities have _ under them)
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  2. #2
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    Seems fairly standard to me. What have you done so far? If you start integrating dx first then it will all work out okay.
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  3. #3
    MHF Contributor Amer's Avatar
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    \int_{1<y<2}\int_{y<x<y^3}-e^{\frac{x}{y}}dxdy

    that is your integral right
    it is the same as

    \int_{1}^{2}\int_{y}^{y^3}-e^{\frac{x}{y}} dx dy

    first we integrate with respect to x and y constant

    \int_{1}^{2}-ye^{\left(\dfrac{x}{y}\right)}\mid_{y}^{y^3}

    sub now

    \int_{1}^{2}-y(e^{\left(\frac{y}{y}\right)}-e^{\left(\frac{y^3}{y}\right)})dy

    \int_{1}^{2}-y(e^{1}-e^{y^2})dy

    \int_{1}^{2}-y(e) dy+ \int_{1}^{2}ye^{y^2} dy

    the first one is easy the second let u=y^2 and you continue
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  4. #4
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    Quote Originally Posted by Amer View Post
    \int_{1<y<2}\int_{y<x<y^3}-e^{\frac{x}{y}}dxdy

    that is your integral right
    it is the same as

    \int_{1}^{2}\int_{y}^{y^3}-e^{\frac{x}{y}} dx dy

    first we integrate with respect to x and y constant

    \int_{1}^{2}-ye^{\left(\dfrac{x}{y}\right)}\mid_{y}^{y^3}

    sub now

    \int_{1}^{2}-y(e^{\left(\frac{y}{y}\right)}-e^{\left(\frac{y^3}{y}\right)})dy

    \int_{1}^{2}-y(e^{1}-e^{y^2})dy

    \int_{1}^{2}-y(e) dy+ \int_{1}^{2}ye^{y^2} dy

    the first one is easy the second let u=y^2 and you continue

    Edit: deleted

    I got it, thanks!
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  5. #5
    MHF Contributor Amer's Avatar
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    look I integrate not derive it is possible that I make mistakes and typos so feel free

    \int e^{\frac{x}{y}} dx

    \dfrac{e^{\frac{x}{y}}}{\left(\frac{d}{dx}(\frac{x  }{y})\right)}

    as you can see the denominator of this is 1/y

    \dfrac{e^{\frac{x}{y}}}{\left(\frac{1}{y}\right)}

    \frac{a}{\left(\frac{c}{d}\right)}=\frac{ad}{c}

    I think it is clear
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