Which couchy theorim do i need

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i got this formula

[f(b)-f(a)]/[g(b)-g(a) = f'(c)/g'(c)

i looked up in google

http://en.wikipedia.org/wiki/Cauchy_theorem

and i got a list of many couchy theorems

which one do i need for this formula??
 
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Cauchy's mean value theorem
 
i can't find a good explanation for Cauchy's mean value theorem
i can't find a video lecture or animation .
 
Did you look at

http://en.wikipedia.org/wiki/Mean_value_theorem#Cauchy.27s_mean_value_theorem

It's basically a 2-dimensional version of the standard mean value theorem. In 1 dimension, you have a function f(t), pick two a and b, and somewhere in between them the slope of the graph is the same as the slope connecting the endpoints of the graph. Similarly, given a graph (f(t),g(t)) (so this is a curve parametrized by t) you pick two endpoints, and somewhere on the curve in between the endpoints is a point where the tangent slope is the same as the slope of the line connecting the points. There's even a pretty picture on wikipedia demonstrating this.
 
i found it

i need video explanation or
flash animation

??
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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