Results 1 to 5 of 5

Math Help - Definate Integral Help

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    1

    Definate Integral Help

    I feel like the solution to this is painfully obviously, but I just can't seem to get it right... Any help would be greatly appreciated.

    f(x)=\int_{3}^{x^3}{t^2}{dt}

    It wants to know f'(x). Wouldn't that just be x^2 then? Since the derivative cancels out the integral? What is messing me up is the limits, since I am pretty sure they make it something other than x^2.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    let  F(x) = \int^{x}_{3} t^{2}dt

    then  \frac{d}{dx} \Big( \int^{x^{3}}_{3} t^{2} dt \Big) = F'(x^{3}) \frac{d}{dx} x^{3} = x^{6} \cdot 3x^{2} = 3x^{8}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    In general, if you have  \int^{v(x)}_{u(x)}f(t)dt

     \frac{d}{dx} \int^{v(x)}_{u(x)}f(t)dt = F'\big(v(x)\big) \frac{d}{dx} v(x) - F'\big(u(x)\big) \frac{d}{dx} u(x)

    where F(x) is an antiderivative of f(t)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member eXist's Avatar
    Joined
    Aug 2009
    Posts
    157
    Random Variable is correct. Sometimes it helps to actually integrate the function, then take the derivative if you need help understanding.

    Ex:

    f(x) = \int_3^{x^3} t^2 \, dt = \frac{x^9}{3} - \frac{27}{3}

    From there, take the derivative:

    \frac{d}{dx}\bigg(\frac{x^9}{3} - \frac{27}{3}\bigg) = \frac{9x^8}{3} - 0 = 3x^8
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    Quote Originally Posted by JayGatsby View Post
    I feel like the solution to this is painfully obviously, but I just can't seem to get it right... Any help would be greatly appreciated.

    f(x)=\int_{3}^{x^3}{t^2}{dt}

    It wants to know f'(x). Wouldn't that just be x^2 then? Since the derivative cancels out the integral? What is messing me up is the limits, since I am pretty sure they make it something other than x^2.
    What is true is that the derivative of \int_a^x f(t) dt is f(x). If you have F(x)= \int_a^{g(x)} f(t)dt, for some function g(x), make a "change of variable". Let u= g(x) so that this becomes F(u)= \int_a^u f(t)dt. Now we have \frac{dF}{dx}= \frac{dF}{du}\frac{du}{dx}. Since u= g(x), \frac{du}{dx}= \frac{dg}{dx} and so \frac{dF}{dx}= f(g(x))\frac{dg}{dx}

    Here, "f(t)" is t^2 and g(x) is x^3 so the derivative is (x^3)^2(3x^2)= 3x^8.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Determine the value of the definate integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 10th 2011, 02:02 AM
  2. definate integral of 'e'
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2009, 04:48 PM
  3. definate integral question
    Posted in the Calculus Forum
    Replies: 6
    Last Post: August 19th 2009, 08:57 AM
  4. Definate Integral help please
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 20th 2009, 01:18 PM
  5. How is it right? (definate integral)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 22nd 2008, 06:32 PM

Search Tags


/mathhelpforum @mathhelpforum