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Differentiation under the integral sign

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data

    I am asked to compute d/dt of [tex]_{c}\int^{d}[/tex] ( [tex]_{a}\int^{t}[/tex] f(x,y)dx)dy for t [tex]\in[/tex] (a,b) for a problem involving differentiation under the integral sign.

    [a,b] , [c,d] are in closed intervals in [tex]\Re[/tex]
    f a continuous real valued function on [a,b] x [c,d]

    2. Relevant equations



    3. The attempt at a solution

    I'm stuck because I don't really get what compute d/dt means in this context. Can someone explain to me what this problem is asking?
     
  2. jcsd
  3. Mar 29, 2009 #2
    Hi celtics2004!

    The double integral defines a real-valued function of t, which turns out to be differentiable. So you can compute its derivative d/dt.
    To compute the derivative, you will need http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign" [Broken] (as you mentioned) and the fundamental theorem of calculus.
     
    Last edited by a moderator: May 4, 2017
  4. Mar 29, 2009 #3

    HallsofIvy

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    Yes, it is asking you to find the derivative of that function!
    [tex]\int_a^t f(x,y)dx[/tex]
    is a function of both t and y. I'm going to call that g(t,y).
    [tex]\int_c^d\int_a^t f(x,y)dx dy= \int_c^d g(t,y)dt[/tex]
    is a function of t only, say h(t). You are being asked to differentiate that function: dh/dt.

    You can use the "fundamental theorem of Calculus", that
    [tex]\frac{d}{dx}\int_a^x f(u)du= f(x)[/itex].

    You can also use the fact that when you are integrating with respect to another variable, you can take the derivative inside the integral:
    [tex]\frac{d}{dx}\int_a^b f(x,y)dy= \int_a^b \frac{\partial f(x,y)}{\partial x} dy[/tex]
     
  5. Mar 29, 2009 #4
    Thanks for your replies. I think I may have figured it out but I'm not very confident in my solution as this subject is still a little confusing.

    Let F(t,y) = [tex]\int^{d}_{c}[/tex] ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.
    Then d/dt F(t,y) = d/dt [tex]\int^{d}_{c}[/tex] ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.
    Using differentiation under the integral sign I get:
    d/dt F(t,y) = [tex]\int^{d}_{c}[/tex] d/dt ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.
    Using the Fundamental Theorem of Calculus I get:
    d/dt F(t,y) = [tex]\int^{d}_{c}[/tex] f(t,y) dy.
    Again using the Fundamental Theorem of Calculus I get:
    d/dt F(t,y) = F(t,d) - F(t,c).

    Does this make sense or did I screw up?

    Thanks again for your help
     
  6. Mar 29, 2009 #5
    The RHS not a function of y, since you already integrated y from d to c. It should just be F(t).

    F is not the antiderivative of f with respect to y in this case, so the last step is not correct. The final answer is what you had before that (without the y on the LHS).
     
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