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!

Fundamental theorem of Calculus

  1. Mar 17, 2014 #1
    Suppose [itex]a[/itex] is a constant.

    If [itex]G(x) = \int_a^x \Big [ f(t) \int_t^x g(u) du \Big ] dt[/itex], what is [itex]G\,'(x)[/itex]?

    My attempt,
    [tex]G\,'(x) = f(x) \int_x^x g(u) du = 0, [/tex]

    and I am sure this is wrong.
     
    Last edited: Mar 18, 2014
  2. jcsd
  3. Mar 18, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you had
    [tex] G(x) = \int_a^x F(x,t) \, dt [/tex]
    in general, what would be your answer for ##G'(x)##? What happens if you specialize this to
    [tex] F(x,t) = f(t) \int_t^x g(u) \, du ? [/tex]
     
  4. Mar 20, 2014 #3
    Thank you for the hint. This is my attempt at the proof which I mimic from that of the single variable case. Please check the rigour in the proof.

    Let
    \begin{equation*}
    G(x) = \int_a^x F(x,t) dt.
    \end{equation*}

    Now
    \begin{align*}
    G(x+h) - G(x)
    &= \int_a^{x+h} F(x+h,t) dt - \int_a^x F(x,t) dt \\[3mm]
    &= \int_a^x F(x+h,t) dt + \int_x^{x+h} F(x+h,t) dt - \int_a^x F(x,t) dt \\[3mm]
    &= \int_a^x F(x+h,t) dt - \int_a^x F(x,t) dt + \int_x^{x+h} F(x+h,t) dt \\[3mm]
    &= \int_a^x \big [ F(x+h,t) - F(x,t) \big ] \; dt + h F(x+h,c) \\[3mm]
    \end{align*}

    for some [itex] c \in (x, x+h) [/itex] by the mean value theorem for integrals. Therefore
    \begin{equation*}
    \frac{G(x+h) - G(x)}{h}
    = \int_a^x \Big [ \frac{F(x+h,t) - F(x,t)}{h} \Big ] \; dt + F(x+h,c) \\[3mm]
    \end{equation*}
    and
    \begin{align*}
    G'(x) &= \lim_{h \to 0} \int_a^x \Big [ \frac{F(x+h,t) - F(x,t)}{h} \Big ] \; dt
    + \lim_{h \to 0} F(x+h,c) \\[3mm]
    &= \int_a^x \frac{\partial}{\partial x} F(x,t) dt + F(x,x)
    \end{align*}
    since as [itex] h \to 0, x + h \to x [/itex] and [itex] c \to x [/itex].

    Now if [itex] F(x,t) = f(t) \int_t^x g(u) du [/itex]. Then

    \begin{equation*}
    \frac{\partial}{\partial x} F(x,t) = f(t) g(x)
    \end{equation*}
    and
    \begin{equation*}
    G'(x) = \int_a^x f(t) g(x) dt + f(t) \int_x^x g(u) du = \int_a^x f(t) g(x) dt.
    \end{equation*}
     
  5. Mar 20, 2014 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    That is correct. There is a general formula for such problems that is often found in good calculus textbooks. Assuming everything is differentiable, we have:
    [tex] \frac{d}{dx} \int_{a(x)}^{b(x)} F(x,t) \, dt = b'(x) F(x,b(x)) - a'(x) F(x,a(x))
    + \int_{a(x)}^{b(x)} F_x(x,t) \, dt [/tex]

    The proof is essentially the same as what you gave.
     
  6. Mar 20, 2014 #5
    I would like to go back a bit and look at:

    If [itex] \quad {\displaystyle G(x) = \int_a^{\psi(x)} f(t) dt } \quad [/itex] then [itex] \quad G\,' (x) = \psi\,' (x) f(\psi(x)). [/itex]

    My attempt at the proof, which again mimic the proof of the basic case:

    Proof

    \begin{align*}
    G(x+h) - G(x)
    &= \int_a^{\psi(x+h)} f(t) dt - \int_a^{\psi(x)} f(t) dt \\[3mm]
    &= \int_{\psi(x)}^{\psi(x+h)} f(t) dt \\[3mm]
    &= \big [\psi(x+h) - \psi(x) \big ] f(\psi(c))
    \end{align*}
    for some [itex] c \in (x, x+h) [/itex] by the mean value theorem.

    Therefore
    \begin{equation*}
    \frac{G(x+h) - G(x)}{h}
    = \frac{\psi(x+h) - \psi(x)}{h} f(\psi(c))
    \end{equation*}
    and
    \begin{align*}
    G\,'(x) &= \lim_{h\to 0} \frac{\psi(x+h) - \psi(x)}{h} f(\psi(c)) \\[3mm]
    &= \psi'(x) f(\psi(x))
    \end{align*}
    since as [itex] h \to 0, x + h \to x [/itex] and [itex] c \to x [/itex].




    The bit that I am not comfortable with is going from

    [itex] {\displaystyle \int_{\psi(x)}^{\psi(x+h)} f(t)} dt \quad [/itex] to [itex] \quad {\displaystyle \big [\psi(x+h) - \psi(x) \big ] f(\psi(c)) } [/itex]

    Perhaps more explanation is needed here to make it rigorous?

    Thank you sir for stating the general case. I saw that in a textbook a long time ago but I did not jot it down and I have been trying to find it ever since. All the Calculus textbooks that I refer to do not state the result.
     
  7. Mar 20, 2014 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The formula Ray is talking about is quoted here http://en.wikipedia.org/wiki/Leibniz_integral_rule. And for justifying your step, you gave the reason. It's a formulation of the mean value theorem. You may need to state some assumptions about continuity, but other than that it's perfectly rigorous.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fundamental theorem of Calculus
Loading...