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Partial derivative of a multivariable integral?

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Stumped.

    Integral: f(x,y) = ∫ (from 1 to xy) of e^(t^2)dt
    find both fx and fy

    3. The attempt at a solution

    I've come up with:
    fx(x,y)
    = /∂x ∫ (from 1 to xy) of e^(t^2)dt

    Not sure where to go... possibly take the integral, the take the partial derivative? I found a solution that moves the derivative into the integral. It just looks like magic to me... not sure what they did. Could someone please break it down for me?

    attachment.php?attachmentid=40270&d=1319385838.jpg
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2011 #2

    HallsofIvy

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    Leibniz's formula:
    [tex]\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt= \frac{db}{dx}f(x,b(x))- \frac{da}{dx}f(x,a(x))+ \int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}dt[/tex]
    For each partial derivative, treat the other variable as a constant of course.
    Here, a(x)= 1, b(x)= xy, and [itex]f(x,y)= e^{t^2}[/itex] which does not depend on x or y so that last integral will be 0.

    ([itex]e^{t^2}[/itex] does not have an elementary integral so that is NOT the way to go!)
     
  4. Oct 23, 2011 #3
    I remember from a problem I did a while back that the integral of e^{t^2} is not the way to go.

    Here's my attempt at using the Liebnitz formula, but my answer is quite different, so perhaps I'm not using it correctly...

    [tex]\frac{d}{dx}\int_{1}^{xy} e^{t^2}dt= \frac{d(xy)}{dx}f(x,xy) - \frac{d(1)}{dx}f(x,1)+ 0[/tex]
    [tex]\frac{d}{dx}\int_{1}^{xy} e^{t^2}dt= \frac{d(xy)}{dx}e^{(xy)^2} - \frac{d(1)}{dx}e^{1^2}[/tex]
    [tex]\frac{d}{dx}\int_{1}^{xy} e^{t^2}dt= 2xye^{(xy)^2} - 2(1)e[/tex]
    [tex]\frac{d}{dx}\int_{1}^{xy} e^{t^2}dt= 2xye^{x^2y^2} - 2e[/tex]

    I think my error might be in where I use say f(x,b(x)→f(x,xy)... I'm not sure how to do this. I just substituted xy for t, but that path doesn't seem to utilize the x component, just the t=xy component of the function. Not even sure if that makes sense.
     
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