How to prove that $\int_0^1 f(x)\,dx = f(0) + \frac{1}{2}f'(c)$ for some $ c \in [0,1]$? 
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*Given that ${f}$ is differentiable on the interval
$[0,1]$ I need to prove that $\int_0^1 f(x)dx = f(0) + \frac{1}{2} f'(c)$ for some $ c \in [0,1]$.

*I'm aware of integral mean value theorem, which gives us the following:

Exists point $c \in [0,1]: {f}(c) = \frac{1}{1} \int_0^1 f(x)\,\mathrm{d}x$
But I can't get further and it looks like that's not the right way at all. I'll be happy to get any tips or key statements that will lead me to the solution, please.

Let $\;a=\int_0^1 f(x)\,dx-f(0)$.
It is sufficient to prove that there exists $c\in\left]0,1\right[$ such that $a=\frac{1}{2}f’(c)$.
We define the following function:
$$\phi(t)=\int_0^t f(x)\,dx+(1-t)f(t)+a(1-t)^2:[0,1]\to\mathbb{R}$$
$\phi(t)$ is differentiable on $[0,1]$ and
$$\phi'(t)=f(t)-f(t)+(1-t)f’(t)-2a(1-t)=(1-t) \left[f’(t)-2a\right]$$
for all $t\in[0,1]$.
$$\phi(0)=f(0)+a=\int_0^1 f(x)\,dx.$$
$$\phi(1)=\int_0^1 f(x)\,dx.$$
Since $\phi(t)$ is a differentiable function on $[0,1]$ and $\phi(0)=\phi(1)$, we can apply Rolle’s Theorem, hence there exists $c\in\left]0,1\right[$ such that $\phi’(c)=0$.
Consequently we get that
$$(1-c)\left[f’(c)-2a\right]=0$$
but $\;1-c>0,\;$ therefore $\;f’(c)-2a=0\;$ that is
$$a=\frac{1}{2}f’(c).$$
