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Homework Help: Derivative of an Integral

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of the function
    [itex]F(x) = \int^0_{x^2-1}\frac{sin(t+1)}{t+1}dt[/itex]


    2. Relevant equations



    3. The attempt at a solution
    [itex]F'(x) = -\frac{sin(x^2)}{x^2}[/itex]

    I'm just learning this and unsure if this is correct. It seems too easy?
     
  2. jcsd
  3. Oct 18, 2011 #2

    Mark44

    Staff: Mentor

    Right. It's not as easy as you are making it. You need to use the chain rule.

    [tex]F(x) = \int^0_{x^2-1}\frac{sin(t+1)}{t+1}dt = -\int_0^{x^2-1}\frac{sin(t+1)}{t+1}dt [/tex]

    The Fundamental Theorem of Calculus says that, if
    [tex]F(x) = \int_0^x f(t)dt [/tex]
    then F'(x) = f(x)

    Notice however, that one of your integration limits is not x, but is instead a function of x.

    [tex]\frac{d}{dx}\int_0^{u} f(t)dt = \frac{d}{du}\int_0^u f(t)dt \cdot \frac{du}{dx}[/tex]

    Now the integral matches the form in the FTC.
     
  4. Oct 18, 2011 #3
    So the answer would be,

    [itex]-\frac{sin(x^2)}{x^2} \cdot 2x[/itex]

    Is this now correct?
     
  5. Oct 18, 2011 #4

    SammyS

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    \cdot for center dot.
     
  6. Oct 18, 2011 #5

    Mark44

    Staff: Mentor

    Looks good, but can be simplified a bit.
     
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