How prove this $\int_{a}^{b}f(x)dx=\frac{1}{2}(b-a)[f(a)+f(b)]-\frac{1}{12}(b-a)^3f''(\xi)$

Let $f(x)$ be a twice-differentiable function on $(a,b)$,show that

there exsit $\xi\in(a,b)$ ,such $$\int_{a}^{b}f(x)dx=\dfrac{1}{2}(b-a)[f(a)+f(b)]-\dfrac{1}{12}(b-a)^3f''(\xi)$$

if this problem condition is Amuss that $f(x)$ is a three-differentiable function,

$$f(x)=f(a)+f'(a)(x-a)+\frac{f''(\xi_{1})\cdot (x-a)^2}{2}$$ $$f(x)=f(b)+f'(b)(x-b)+\frac{f''(\xi_{2})\cdot (x-b)^2}{2}$$ so $(1)+(2)$ $$\Longrightarrow 2f(x)=f(a)+f(b)+f'(a)(x-a)+f'(b)(x-b)+\dfrac{1}{2}[f''(\xi_{1})+f''(\xi_{2})][(x-a)^2+(x-b)^2]$$ since $f(x)$ is a three-differentiable function,so $f''(x)$ is continuous,so there exsit $\xi\in(\xi_{1},\xi_{2})$,such $$\dfrac{1}{2}[f''(\xi_{1})+f''(\xi_{2})]=f''(\xi)$$ $$\int_{a}^{b}f(x)dx=\dfrac{1}{2}(b-a)[f(a)+f(b)]-\dfrac{1}{12}(b-a)^3f''(\xi)$$ But if $f(x)$ have twice-differentiable,this methods is not usefull

Then I use this methods can't prove it.

Since $(b-a)\frac{f(a)+f(b)}{2}$ is the integral over $[a,b]$ of the linear function whose values in $a,b$ are $f(a),f(b)$, we just need to show that for any twice differentiable function over $[0,1]$, such that $f(0)=f(1)=0$, $$\int_{0}^{1}f(x)\,dx = -\frac{1}{12} f''(\xi),\qquad \xi\in(0,1).\tag{1}$$ By exploiting the condition $f(0)=f(1)=0$ and using integration by parts we have:

$$\int_{0}^{1}f(x)\,dx = -\int_{0}^{1}x\,f'(x)\,dx = \int_{0}^{1}\left(\frac{1}{2}-x\right)\,f'(x)\,dx=\int_{0}^{1}\frac{x^2-x}{2}\,f''(x)\,dx\tag{2}$$ hence the claim follows by applying the mean value theorem in the (weighted) form: $$\int_{a}^{b}u(x) v(x)\,dx = u(\xi)\int_{a}^{b}v(x)\,dx$$ holding for $v(x)=\frac{x-x^2}{2}\geq 0$ and $u(x)=f''(x)$.

Quoting china math

"since $f(x)$ is a three times differentiable function hence $f''(x)$ is continuous, so there exist $\xi\in(\xi_{1},\xi_{2})$, such that $$\dfrac{1}{2}[f''(\xi_{1})+f''(\xi_{2})]=f''(\xi) \tag{1}$$"

This is also true for a twice differentiable function!

Darboux theorem says that let $I$ be an open interval and $f:I\rightarrow \mathbb{R}$ a real valued function. Then $f'$ has the intermediate value property.

In your proof, modify your assumption ($f$ is a twice derivable function) and complete the proof using Darboux theorem and you have one nice proof. By the way, there is a similar problem I have answered previously here (there is a hint on how to invoke Darboux theorem to prove $(1)$).