How can I prove that there exists $c\in[0,1]$ such that $\int\limits_0^1\int\limits_0^tf''(x)dxdt=\frac{1}{2}f''(c)$ How can I prove that there exists $c\in[0,1]$ such that $$\int\limits_0^1\int\limits_0^tf''(x)dxdt=\frac{1}{2}f''(c)$$
I used the MVT so
$$\int\limits_0^1\int\limits_0^tf''(x)dxdt=\int\limits_0^{c_1}f''(x)dx$$ for some $c_1\in[0,1]$ and applying it again $$\int\limits_0^{c_1}f''(x)dx=c_1f''(c_2)$$ for some $c_2\in[0,c_1]$, but that is not what I want. Tanks.

This is just the error term when you Taylor expand the function $f$ at 0. First of all, the definite integral is equivalent to $f(1)-f(0)-f'(0)$. General fact of Taylor series tells us that $f(x)=f(0)+\frac{f'(0)}{1!}(x-0) + \frac{f^{''}(c)}{2!}(x-0)^2$ for some $c\in [0,1]$. Plug in x=1 done. You may want to verify that fact.
