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This is a continuation of the question asked in second-order divided differenc.

Suppose the periodic function $f(t)$ has only one value, $f(t_0) = -10$ within the period $[0,T]$, all other values of the function being zero within that period. We can interpolate $f(t)$ by $asin(t) - 10$ which for $t_0 = 0$ will give the value $-10$. Now, if we accept the interpolation then $f'(t) = acos(t)$ and therefore for $t_0=0$ we will have $f'(t_0) = acos(0) = a$. Thus, at $t_0 =0$ we will obtain $f(t_0)f'(t_0) = -10a$. At the end of the period the value of $f(t)f'(t)$ will be also $-10a$. Therefore, the average value of $f(t)f'(t)$ for the period is $\frac{2(-10a)}{2} = -10a$. Do you agree with this?

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If I understand your question correctly, you have an $f: [0,T] \to \mathbb{R}$, $ f(t) = \begin{cases} -10 &\text{if } t=t_0 \\ 0 &\text{otherwise} \end{cases} $ and are asking if for the value of $ \frac{1}{T}\int_0^T f(t)f'(t) dt \text{ .} $

That question doesn't make a lot of sense then, because $f$ is not differentable, hence $f'(t)$ isn't even defined. And if you go on to define $f'(t)$ somehow, the integral will be $0$, because the integrand is zero everywhere except at $t_0$.

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    Since the times civilized discussions are established in the modern world -- a couple of centuries or so ago. Down-voting and disappearing is an easy and mean way out when one feels that he/she should win the discussion at any rate.2012-10-20