# fundamental theorem of calculus relation

I sent this question last week but it turns out that I had an error in the formula that I was trying to understand. So, here goes a second try. I am reading a proof in a convex optimization book by Nesterov and at the beginning of the proof it says the following:

for any $x,y \in \mathbb{R}^n$, we have

$f'(y) = f'(x) + \int_0^1 f''\left(x + \tau(y-x)\right)(y-x)~d\tau$

The earlier answer referred to the fundamental theorem of calculus but I read about the fundamental theorem of calculus and I'm still not seeing it. Thanks a lot.

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I think you want $f'(y)-f'(x)$, not $f'(y) = f'(x)$. – Muphrid Jan 8 '13 at 17:23
thanks muphrid: I fixed it. – mark leeds Jan 8 '13 at 18:01

Consider the function $$g(\tau):=f'(x+\tau (y-x)).$$ Then by chain rule $$g'(\tau)=f''(x+\tau (y-x))(y-x).$$ Now the fundamental theorem of calculus yields $$g(1)-g(0)=\int_0^1 g'(\tau)d\tau$$ which can be rewritten $$f'(y)-f'(x)=\int_0^1 f''(x+\tau (y-x))(y-x)d\tau.$$

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Let $s = \tau(y-x) + x$. Then $ds = (y-x)d\tau$ or $d\tau = ds/(y-x)$. As $\tau$ runs from $[0,1]$, $s$ runs from $[x,y]$. The integral then becomes

$$\int_x^y f''(s) (y-x) \frac{ds}{y-x} = \int_x^y f''(s) \, ds = f'(y) - f'(x)$$

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thanks to julien and muphrid. I'll print these out and see if I can follow. they are much appreciated. this site is amazing. – mark leeds Jan 8 '13 at 18:01