Noether's theorem time invariance -- mean value theorem use?

[Tush19]

how does the first step use mean value theorem? I don't get it , can anyone explain , thanks.

 

[Orodruin]

The mean value theorem states that
$$
\int_x^{x+\delta x} f(s) ds = \delta x\, f(x^*)
$$
where $x \leq x^* \leq x + \delta x$. Since $f$ is continuous, $f(x^*) \to f(x)$ for small $\delta x$.

 

[Tush19]

The mean value theorem states that
$$
\int_x^{x+\delta x} f(s) ds = \delta x\, f(x^*)
$$
where $x \leq x^* \leq x + \delta x$. Since $f$ is continuous, $f(x^*) \to f(x)$ for small $\delta x$.

thanks but I couldn't find that mean value theorem statement anywhere ,all it shows that mean value theorem is the following

 

[DrClaude]

thanks but I couldn't find that mean value theorem statement anywhere ,all it shows that mean value theorem is the following

https://www.kristakingmath.com/blog/mean-value-theorem-for-integrals

 

[Orodruin]

thanks but I couldn't find that mean value theorem statement anywhere ,all it shows that mean value theorem is the following
View attachment 295180

Wrong mean value theorem:
https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_definite_integrals

 

[Orodruin]

Just to add: The mean value theorem for definite integrals is easy to obtain from the theorem you quoted. Just consider that
$$
(b-a) f’(c) = f(b) - f(a) = \int_a^b f’(x) dx
$$
and let $g(x) = f’(x)$. You now have
$$
\int_a^b g(x) dx = (b-a) g(c)
$$
for some $c$ such that $a\leq c\leq b$.

 

[Tush19]

Just to add: The mean value theorem for definite integrals is easy to obtain from the theorem you quoted. Just consider that
$$
(b-a) f’(c) = f(b) - f(a) = \int_a^b f’(x) dx
$$
and let $g(x) = f’(x)$. You now have
$$
\int_a^b g(x) dx = (b-a) g(c)
$$
for some $c$ such that $a\leq c\leq b$.

thank you so much