In summary: for clearing that up for me ,you're a lifesaverthank you so much for clearing that up for me ,you're a lifesaver
  • #1
Tush19
3
0
Screenshot 2022-01-06 at 00.54.20.png

how does the first step use mean value theorem? I don't get it , can anyone explain , thanks.
 
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  • #2
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##.
 
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  • #3
Orodruin said:
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
1641479493425-png.png
 
  • #6
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##.
 
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  • #7
Orodruin said:
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
 

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