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I am trying to simplify Leibniz Rule to the (first) Fundamental Theorem of Calculus (FTC) but believe I am doing so incorrectly. Leibniz rule can be written as:

$$\frac{d}{dt} \int_{f(t)}^{g(t)} A(t,\sigma) d\sigma = A(t,g(t))\dot g(t) - A(t,f(t))\dot f(t) + \int_{f(t)}^{g(t)} \frac{\partial}{\partial t} A(t,\sigma) d\sigma \qquad (1)$$

If I set $f(t)=c=const$ and $g(t)=t$ this simplifies to

$$\frac{d}{dt} \int_{c}^{t} A(t,\sigma) d\sigma = A(t) + \int_{c}^{t} \frac{\partial}{\partial t} A(t,\sigma) d\sigma \qquad (2)$$

Now if I assume $A$ does not depend on $t$ s.t. $A=A(\sigma)$ then

$$\frac{d}{dt} \int_{c}^{t} A(\sigma) d\sigma = A(t) + \int_{c}^{t} \frac{\partial}{\partial t} A(\sigma) d\sigma \qquad (3)$$

which simplifies to

$$\frac{d}{dt} \int_{c}^{t} A(\sigma) d\sigma = A(t) \qquad (4)$$

which is the (first) FTC. But what happens if instead of assuming $A$ does not depend on $t$, we assumed $\sigma=t$? We get

$$\frac{d}{dt} \int_{c}^{t} A(t) dt= A(t) + \int_{c}^{t} \frac{\partial}{\partial t} A(t) dt \qquad (5)$$

which can be proven incorrect by setting $A(t) = t^2$ yeilding

$$t^2=t^2+t^2-c^2=2t^2-c^2 \qquad (6)$$

I don't understand where my error in logic is. Can anyone please help? I'm trying to understand how the (first) FTC applies to functions of time like velocity; i.e. the following should be true

$$\frac{d}{dt} \int_{c}^{t} v(t) dt= v(t) \qquad (7)$$

Please let me know if I need to be more specific or clarify anything. Many thanks.

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    $\begingroup$ It does not make sense to set $t = \sigma$, since $\sigma$ is a "mute" integration variable $\endgroup$ – PierreCarre Feb 20 at 15:28
  • $\begingroup$ Is a "mute" variable the same as a "dummy" variable? If so, I think I see your point, but I'm not clear on how i can get to equation (7), where the integrand depends only on $t$, without making this (nonsensical) assumption. $\endgroup$ – eball Feb 20 at 16:07
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    $\begingroup$ yes, leibnitz works only for the specified form on top of your question. so what you do is just don't use leibnitz rilule but the ftc. just integrate the velocity. you get V(t) - V(c). if you differentiate again with respect to t, what do you get? $\endgroup$ – Luke Feb 20 at 16:32
  • $\begingroup$ @Luke, I thought Leibniz Rule was a more general form of the (first) FTC. So shouldn't you be able to go from Leibniz Rule to the (first) FTC? If I'm not mistaken: $\frac {d (V(t) - V(c))} {dt}=v(t)$ $\endgroup$ – eball Feb 20 at 16:47
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    $\begingroup$ my apologies, sure it works. you can't set $t=\sigma$ because the leibnitz rule makes a statement about integrating with respect to some other variable than t. so it works in your example if you integrate $v(\sigma) d\sigma$. the important thing is that the differential is not taken wrt to the variable of integration. if you change that, you get the wrong thing, as the others already pointed out. of course you can always achieve such a thing. using leibnitz instead of ftc on the velocity integral is like using a sledgehammer to crack a nut, still its possible by setting $ v(t)=v(t, \sigma) $ $\endgroup$ – Luke Feb 20 at 17:20
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I got the answer now. If you have $$\frac{d}{dt} \int_{c}^{t} A(t, \sigma) d\sigma= A(t, t) + \int_{c}^{t} \frac{\partial}{\partial t} A(t,\sigma) d\sigma $$ where $A(t,\sigma)= t^2$ then you get on the LHS (by integrating and differentiating afterwards) and on the RHS (by taking partial derivative and then integrating) the same value, namely $3t^2-2ct$, so the theorem holds in this case.

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  • $\begingroup$ Hey @Luke, I'm having trouble following your work. We agree that on the LHS you get $\frac{d}{dt} \int_{c}^{t} A(t,\sigma) d\sigma=t^2$, but on the RHS shouldn't you get $A(t) + \int_{c}^{t} \frac{\partial}{\partial t} A(t,\sigma) d\sigma=t^2+(t^2-c^2)$? $\endgroup$ – eball Feb 21 at 23:31
  • $\begingroup$ we don't agree on both sides, and that was your mistake actually (and i admit, mine too). I'll explain the LHS and you should then try the RHS. $\frac{d}{dt} \int_{c}^{t} A(t,\sigma) d\sigma = \frac{d}{dt} \int_{c}^{t} t^2 d\sigma = \frac{d}{dt} t^2 \int_{c}^{t} 1 d\sigma = \frac{d}{dt} t^2 (t-c) = \frac{d}{dt} t^3 - t^2 \cdot c = 3t^2-2tc$ $\endgroup$ – Luke Feb 21 at 23:39
  • $\begingroup$ the problem was, of course, that the Leibnitz rule can only be equal to the FTC if you had a function depending only on sigma (which you noticed correctly in (1)-(4)). but if the function depends only on t, or on t and sigma, you can't apply the FTC because thats not how the FTC works. that is exactly why the Leibnitz rule extends the FTC, and i apologize for causing confusion before. It's been a while since I needed that. $\endgroup$ – Luke Feb 21 at 23:54
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    $\begingroup$ I got it to work! Thank you so much for the explanation!! (i truly appreciate you spelling it out haha) I think i get it now $\endgroup$ – eball Feb 22 at 3:12

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