Results 1 to 6 of 6
Like Tree7Thanks
  • 3 Post By Idea
  • 1 Post By jumbo1985
  • 1 Post By IvanM
  • 1 Post By jumbo1985
  • 1 Post By IvanM

Thread: Two definite integrals are equal, what about intervals of integration?

  1. #1
    Newbie
    Joined
    Jan 2016
    From
    USA
    Posts
    14
    Thanks
    2

    Two definite integrals are equal, what about intervals of integration?

    Suppose I have two simple, integrateable functions of variable t such that

    $\displaystyle \int_{0}^{t_1}f(t)dt = \int_{0}^{t_2}g(t)dt$

    I believe that if $\displaystyle f(t) \leq g(t)$ for all t then $\displaystyle t_2 \leq t_1$

    Any tips on how to prove this? Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,039
    Thanks
    530

    Re: Two definite integrals are equal, what about intervals of integration?

    is it given that $f$ and $g$ are non-negative functions?
    Thanks from topsquark, jumbo1985 and IvanM
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2016
    From
    USA
    Posts
    14
    Thanks
    2

    Re: Two definite integrals are equal, what about intervals of integration?

    Sorry, I forgot to mentioned anything about that. Yes, both functions are non-negative.
    Thanks from IvanM
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2019
    From
    New York
    Posts
    12
    Thanks
    3

    Re: Two definite integrals are equal, what about intervals of integration?

    Here is my approach to it, I hope this helps:

    So here you are starting off with the equation,

    $\displaystyle \int_{0}^{t_{1}}f(t)\,dt=\int_{0}^{t_{2}}g(t)\,dt $.

    If $\displaystyle g(t) ≥ f(t) $ for all $\displaystyle t $, then:

    $\displaystyle g(t)=f(t)+m(t) $

    where $\displaystyle m(t) $ is an arbitrary function which makes the two equations above true.

    In case that $\displaystyle g(t)=f(t) $, $\displaystyle m(t)=0 $.

    So now we can try substituting that into your original equation and we get,

    $\displaystyle \int_{0}^{t_{1}}f(t)\,dt=\int_{0}^{t_{2}}f(t)\,dt+ \int_{0}^{t_{2}}m(t)\, dt $.

    Move over the $\displaystyle \int_{0}^{t_{2}}f(t)\, dt $ to the left side and what's left is:

    $\displaystyle \int_{0}^{t_{1}}f(t)\,dt\, -\int_{0}^{t_{2}}f(t)\,dt\,=\int_{0}^{t_{2}}m(t)\, dt $.

    If $\displaystyle f(t) $ and $\displaystyle g(t) $ are non-negative functions and $\displaystyle g(t)≥f(t) $ for all $\displaystyle t $, $\displaystyle m(t) $ has to be a non-negative function as well, and the previous equation is only true when $\displaystyle t_{1} ≥ t_{2} $. If $\displaystyle t_{2} > t_{1} $ (which would therefore make $\displaystyle f(t)≥g(t) $), we would get $\displaystyle -\int_{0}^{t_{2}}m(t)\, dt $. However, $\displaystyle m(x) $ is not a negative function in our case, so its integral cannot be negative if all previous derivations hold true.

    And of course, if $\displaystyle t_{1} = t_{2} $, then:

    $\displaystyle \int_{0}^{t_{1}}f(t)\,dt\, -\int_{0}^{t_{2}}f(t)\,dt\,=0 $.

    Hope that helped, and feel free to ask any questions!
    Thanks from jumbo1985
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2016
    From
    USA
    Posts
    14
    Thanks
    2

    Re: Two definite integrals are equal, what about intervals of integration?

    Thank you Ivan. I like your approach.

    Here's my version. I make an additional assumption that $t_1, t_2 > 0$ which is missing from the original post. If you guys see any flaw with this please let me know. Thanks


    Let $g(t) = f(t) + m(t)$. Then
    \begin{eqnarray}
    \int_{0}^{t_2} g(t)dt = \int_{0}^{t_2}f(t)dt + \int_{0}^{t_2}m(t)dt\nonumber
    \end{eqnarray}

    If $m(t) = 0$ then $t_1 = t_2$.

    Otherwise $m(t) > 0$. Let $t_1 = t_2 + \triangle t$ with $\triangle t \in R$.

    The case $\triangle t = 0$ is not possible as it leads to a contradiction in the form of
    \begin{eqnarray}
    \int_{0}^{t_1}f(t)dt < \int_{0}^{t_1}f(t)dt + \int_{0}^{t_1}m(t)dt
    \end{eqnarray}

    Then $\triangle t < 0 \lor \triangle t > 0$.

    Assume that $\triangle t < 0$. Then
    \begin{eqnarray}
    \int_{0}^{t_2} g(t)dt = \int_{0}^{t_1} g(t)dt + \int_{t_1}^{t_2} g(t)dt\nonumber\\
    = \int_{0}^{t_1}f(t)dt + \int_{0}^{t_1}m(t)dt + \int_{t_1}^{t_2}f(t)dt + \int_{t1}^{t_2}m(t)dt\nonumber\\
    = \int_{0}^{t_1}f(t)dt + M
    \end{eqnarray}
    This is not possible since $M > 0$ and therefore $\triangle t > 0$. $\square$
    Last edited by jumbo1985; Aug 15th 2019 at 07:22 PM.
    Thanks from IvanM
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2019
    From
    New York
    Posts
    12
    Thanks
    3

    Re: Two definite integrals are equal, what about intervals of integration?

    Yeah that's a really good approach to the problem as well, thanks!
    Thanks from jumbo1985
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Apr 7th 2014, 12:52 PM
  2. Replies: 5
    Last Post: Dec 5th 2011, 05:21 PM
  3. Replies: 4
    Last Post: Apr 13th 2011, 02:08 AM
  4. definite integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jul 17th 2010, 12:14 PM
  5. definite integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 11th 2010, 09:34 AM

/mathhelpforum @mathhelpforum