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Calculus 2 second derivative of integral

  1. Jan 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Find d2/dx2 0x (1sint√(1+u^4)du)dt


    2. Relevant equations



    3. The attempt at a solution
    Initially I treated this problem as the second derivative of a double integral and thus quickly found myself at the result cosx√1+sin4x, by the fundamental theorem of calculus. However I realized that the problem asks for the second derivative according to x, and now I think my solution is incorrect. Would the correct method be to solve the integral √(1+u^4)du, then do the derivatives?
     
    Last edited: Jan 25, 2014
  2. jcsd
  3. Jan 25, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    Your solution is correct. If it helps, define
    [tex]
    f(t) = \int_1^{\sin t} \sqrt{1 + u^4}\,du.
    [/tex]
    Then
    [tex]
    \frac{d}{dx} \int_0^x \int_1^{\sin t} \sqrt{1 + u^4}\,du\,dt
    = \frac{d}{dx} \int_0^x f(t) \,dt
    = f(x)
    [/tex]
    and
    [tex]
    \frac{df}{dx} = \frac{d}{dx} \int_1^{\sin x} \sqrt{1 + u^4}\,du
    = \cos x \sqrt{1 + \sin^4 x}
    [/tex]
     
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