# 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.

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 $(1)$).

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 $u(x)=f''(x)$.

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.

• It might also be instructive to use the nascent function $f_\epsilon(x)=\sqrt{\epsilon^2+x^2}$. That way it's twice-differentiable for all $\epsilon>0$ but not for $\epsilon=0$. – Semiclassical Sep 30 '14 at 2:06