1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

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

    Mark44

    Staff: Mentor

    Looks good, but can be simplified a bit.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Derivative of an Integral
Loading...