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Derivative of integral bounded by functions

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data
    What's the derivative of the following two:

    2. Relevant equations

    3. The attempt at a solution
    I thought of doing the following:
    [tex]\int_{h(a)}^{h(x)}f(t)\,\mathrm{d}t = \int_{a}^{x}f\circ h(u)\cdot h'(u)\,\mathrm{d}u[/tex]
    (with t = h(u) )
    But then I don't know how to continue.

    Thanks for your help.
  2. jcsd
  3. Jan 20, 2009 #2


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    If the antiderivative of f(t) is F(t), then the integral of the second one is F(v(x))-F(u(x)), right? Differentiate that using the chain rule.
  4. Jan 20, 2009 #3


    Staff: Mentor

    These problems are applications of the Fundamental Theorem of Calculus. As a hint, I'll work a different, but related, problem.

    [tex]\int_a^{x^2} f(t) dt[/tex]
    Suppose an antiderivative of f(t) is F(t). I.e., F'(t) = f(t).
    [tex]\int_a^{x^2} f(t) dt = F(x^2) - F(a)[/tex]
    If we differentiate both sides of this equation with respect to x, we get
    [tex]d/dx \int_a^{x^2} f(t) dt = d/dx(F(x^2) - F(a))[/tex]
    = F'(x^2) * d/dx(x^2) - 0 (F(a) is a constant, so its derivative is zero)
    = f'(x^2) * 2x

    Is that enough to get you going?
  5. Jan 20, 2009 #4
    OK, so I guess then...
    [tex]\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{u(x)}^{v(x)}f(t)\,\mathrm{d}t \right)=v'(x) \cdot f \circ v(x) - u'(x) \cdot f \circ u(x)[/tex]
    As for the other one, since u'(x) = da/dx = 0, we have:
    [tex]\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{a}^{h(x)}f(t)\,\mathrm{d}t \right)= h'(x)\cdot f\circ h(x)[/tex]

    Is that all OK?
  6. Jan 20, 2009 #5


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    Looks ok to me.
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