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Let $f$ be twice continuously differentiable. Prove that there exists $\xi\in (-1,1)$ such that $$\int_{-1}^1 xf(x)dx=\frac{2}{3}f'(\xi)+\frac{1}{3}\xi f''(\xi).$$

What I have tried is as follows.

$$\frac{2}{3}f'(\xi)+\frac{1}{3}\xi f''(\xi)=\frac{1}{3}[xf(x)]''|_{x=\xi}.$$

But then how can we do...

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2 Answers 2

up vote 2 down vote accepted

Put $g(x)=xf(x)$, $\displaystyle H(x)=\int_{-1}^x g(t)dt$, and $\displaystyle G(x)=\int_{x}^{1} g(t)dt$. The function $H$ is $3$ time differentiable, and $H^{\prime}(x)=g(x)$, $H^{\prime\prime}(x)=g^{\prime}(x)$ and $H^{\prime\prime\prime}(x)=g^{\prime\prime}(x)$.

We have for all $x$ that there exists $c_x\in ]-1,1[$ (depending on $x$) such that:

$$H(x)=H(0)+xH^{\prime}(0)+\frac{x^2}{2}H^{\prime\prime}(0)+\frac{x^3}{6}H^{\prime\prime\prime}(c_x)$$ Hence, if we put $x=1$, there exists $c_{1}\in ]-1,1[$ such that:

$$H(1)=H(0)+\frac{f(0)}{2}+\frac{g^{\prime\prime}(c_1)}{6}$$

We have that $G^{\prime}(x)=-g(x)$. In the same way,

$$G(x)=G(0)+xG^{\prime}(0)+\frac{x^2}{2}G^{\prime\prime}(0)+\frac{x^3}{6}G^{\prime\prime\prime}(d_x)$$

and there exists $d_{-1}\in ]-1,1[$ such that

$$G(-1)=G(0)-\frac{f(0)}{2}+\frac{g^{\prime\prime}(d_{-1})}{6}$$

Now $\displaystyle I=\int_{-1}^1 xf(x)dx=H(1)=G(-1)=H(0)+G(0)$. We get:

$$I=\frac{1}{6}(g^{\prime\prime}(c_1)+g^{\prime\prime}(d_{-1}))$$

Suppose wlog that $d_{-1} \leq c_1$, and put $J=[d_{-1}, c_1]$.

If $\displaystyle M={\rm Sup}\{g^{\prime\prime}(x), x\in J\}$ and $\displaystyle m={\rm Inf}\{g^{\prime\prime}(x), x\in J\}$, we have:

$$m\leq 3I\leq M$$ As $g^{\prime\prime}$ is continuous, there exists $c\in J\subset (-1,1)$ such that $g^{\prime\prime}(c)=3I$, and we are done.

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Lemma. Let $g(x)$ is twice continuously differentiable. Then there exists $\xi\in(-1,1)$ such that $$\int\limits_{-1}^1 g(x)\,dx=2g(0)+\frac13 g''(\xi).$$

Proof: Expanding $g(x)$ by Taylor formula we get $g(x)=g(0)+xg'(0)+\frac{x^2}{2}g''(\xi(x))$, where $\xi(x)\in(-1,1)$. Then $$\int\limits_{-1}^1 g(x)\,dx=g(0)\int\limits_{-1}^1\,dx+g'(0)\int\limits_{-1}^1x\,dx+ \frac12\int\limits_{-1}^1x^2g''(\xi(x))\,dx=2g(0)+\frac12\int\limits_{-1}^1x^2g''(\xi(x))\,dx.$$ By the first mean value theorem for integration there exists $x_1\in(-1;1)$ such that $$\int\limits_{-1}^1x^2g''(\xi(x))\,dx=g''(\xi(x_1))\int\limits_{-1}^1x^2\,dx=\frac23g''(\xi(x_1)),$$ since $x^2\geq 0$ and $g''(\xi(x))=\cases{\dfrac{2(g(x)-g(0)-xg'(0))}{x^2},\quad x\neq0 \\ g''(0),\quad x=0}$ is a continuous function. Denoting $\xi=\xi(x_1)$ we obtain the required formula.

To get the result in question we just need to take in this lemma $g(x)=xf(x)$.

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2  
I believe $\xi$ depends on $x$, no? –  user1337 Jul 7 at 9:28
1  
YOu cannot factor $g''(\xi)$ out from the integral, as $\xi=\xi(x)$! –  daw Jul 7 at 9:28
    
@daw, user1337, thank you! You're right. I edited the solution, now this is taken into account. –  CuriousGuest Jul 7 at 12:11

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