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Basic question on double integrals

  1. Jul 8, 2010 #1
    sorry for the trivial question: what's the correct way of computing the following double integral:

    [tex]\int_a^b \int_c^d \frac{\partial^2 f}{\partial x \partial y} dy dx[/tex]
  2. jcsd
  3. Jul 8, 2010 #2


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    Assuming things are reasonably well behaved, the answer is f(b,d)-f(a,d)-f(b,c)+f(a,c), where f is f(x,y).
  4. Jul 8, 2010 #3


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    Since the differentials are ordered "dy dx", this means:
    [tex]\int_a^b\left(\int_c^d \frac{\partial^2 f}{\partial x\partial y} dy\right)dx[/tex]

    (I would consider it better to write
    [tex]\int_{x= a}^b\int_{y= c}^d \frac{\partial^2 f}{\partial x \partial y} dy dx[/tex])

    By the fundamental theorem of of calculus,
    [tex]\int_c^d \frac{\partial^2 f}{partial y\partial x} dy= \frac{df(x,d)}{dx}- \frac{df(x,c)}{dx}[/tex]
    so that
    [tex]\int_{x=a}^b\int_{y= c}^d \frac{\partial^2 f}{\partial x\partial y}dy dx= \int_a^b \left(\frac{df(x,d)}{dx}- \frac{df(x,c)}{dx}\right)dx[/tex]

    Applying the fundamental theorem of calculus again gives mathman's solution.
  5. Jul 9, 2010 #4
    thank you both.
    very clear answers.
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