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.
Best Answer
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.