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.
 A: 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.
A: Let's integrate.
$$
\int_0^t f''(x)\,dx=f'(t)-f'(0)\\
\int_0^1 \int_0^t f''(x)\,dx\,dt=\int_0^1 [f'(t)-f'(0)]\,dt = f(1)-f(0)-f'(0)
$$
By Taylor
$$
f(x)=f(0)+f'(0)x + \frac12f''(c)x^2, \quad c\in[0,1].
$$
Substitute $x=1$ and get result.
A: We know that if $g$ is continuous on $[a,b],$ then $\int_a^b g(t)\,dt = g(c)(b-a)$ for some $c\in [a,b].$ Similarly, if $g$ is continuous on the triangle $T = \{(x,t): 0\le x \le t, t\in [0,1]\},$ then
$$\int\int_T g(x,t)\,dx\,dt = g(c,d)\cdot A(T)$$
for some $(c,d)\in T,$ where $A$ denotes area. (The proof is the same as in one variable; the connectivity of $T$ is the key.) In the problem at hand we can take $g(x,t) = f''(x)$ and the answer falls right out.
