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

• A
Tush19

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

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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 and vanhees71
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

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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##.

DrClaude, Tush19 and vanhees71
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