Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

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


    Staff: Mentor

    Looks good, but can be simplified a bit.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook