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.
 A: 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 $(1)$).
A: 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 $u(x)=f''(x)$.
A: For fun, I plugged in $f(x)=|x|$ and $a=-1, b=1$, which does not have second derivative at $x=0$. 
$$ 1 = \int_{-1}^1 |x| \; dx = \tfrac{1}{2} \big(1 - (-1)\big) \big[ f(1)+f(-1)\big] - \tfrac{1}{12}\big(1 - (-1)\big)\, f''(\xi)$$
After simplification, the linear term vanishes and you get the strange relation that 
$$ 1 =  2 - \tfrac{1}{6} f''(\xi)$$
for some $-1 < \xi < 1$.  Presumably $\boxed{\xi=0}$  The result is instructive.

Working out @Semiclassical's example.  Let $x = \epsilon \tan \theta$ and $t = \sin \theta$.
$$ \int_{-1}^1 dx \; \sqrt{\epsilon^2 + x^2 } 
= \underbrace{\int_{-1+\epsilon}^{1-\epsilon} dx \; \sqrt{\epsilon^2 + x^2 }}_{I}
+ \underbrace{\int_{-\epsilon}^{+\epsilon} dx \; \sqrt{\epsilon^2 + x^2 }}_{II}
$$
Let's estimate I and II. 
$$ \int_{-1+\epsilon}^{1-\epsilon} dx \; \sqrt{\epsilon^2 + x^2 }
\approx \int_{-1+\epsilon}^{1-\epsilon} dx \; |x|
= 1 - \epsilon^2 $$
What really matters is around $x=0$.  
$$  \int_{-\epsilon}^{+\epsilon} dx \; \sqrt{\epsilon^2 + x^2 } 
\approx \int_{-\epsilon}^{+\epsilon} dx \; \epsilon\sqrt{1 + (x/\epsilon)^2 }
\approx  \int_{-\epsilon}^{+\epsilon} dx \; \epsilon\left(1 + \frac{x^2}{2\epsilon^2}\right) = 2\epsilon^2 + \frac{\epsilon}{6}$$
This was a no-frills computation may irritate some analysts.  The idea is to get a convincing answer quickly.  And it even looks like a local version of the 1st part above. 

Or try a Fourier series, $f(\theta) = \sum c_n e^{2\pi i n \theta}$.  Then 
$$ \int_0^1 f(x) dx = 0 = \frac{1}{12} f''(\xi)$$
So there is at least 1 (in fact, many) point(s) of inflection $0 < \xi < 1$ for any Fourier series.
