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.

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.
