Deriving the Average Value of a Function using the Mean Value Theorem

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Homework Statement



If fave [a,b] denotes the average value of f on the interval [a,b] and a<c<b, show that
fave[a,b] = (c-a)/(b-a) fave[a,c] + (b-c)(b-a) fave[c,b]

Homework Equations



All i know is the mean value theorem for integrals is f(c) = fave = 1/(b-a) integral(f(x),x,b,a)

The Attempt at a Solution



Tried using the theorem, but had no idea how to get to that point.

Thanks!
 
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Hi,

Why don't you apply the definition of the fave to fave[a,c] and fave[c,b]?

Once you do this, I think you'll see that your expression simplifies quite nicely.
 
Hello, I tried it and i believe this is how it goes...

fave[a,c] = 1/(c-a) [f(c)-f(a)]
fave[c,b] = 1/(b-c) [f(b)-f(c)]

then i add it together or what? Kind of confused on what to do because this gives something weird...

I do know that u can split up the bounds from [a,c] and [c,b] to get [a,b].. does that correlate with what this got to do?
 
Perhaps I don't understand your notation, but shouldn't [f(c)-f(a)] be Int[f, a, c]?

Try plugging in those expressions into the right-side of the equation that you're trying to prove.
 
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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