Let $f(x)$ is continuous function having continuous 2nd derivative on $[0,1]$

Question

Let $f(x)$ is continuous function having continuous 2nd derivative on $[0,1]$ . Show that there is a point $ \xi \in (0,1) $ such that $$ \int_{0}^{1} xf(x)dx=\frac{1}{2} f(2/3)+ \frac{1}{72} f''(\xi) . $$

I think we can apply Second mean value theorem , Bonnet's form. Here , $f(x)$ continuous on $[0,1]$ and hence it is bounded there and $x$ is non-decreasing on $[0,1]$ , hence there exists $ \xi \ \in (0,1) $ such that $ \displaystyle\int_{0}^{1} xf(x)dx= f(\xi) \int_{\xi}^{1} xdx+ f(\xi) \int_{0}^{\xi} xdx =\frac{1}{2} f(\xi)-1/2 *\xi^{2}/2 $. But I can not proceed further. please help me

Answer 1

Use the second-order Taylor expansion about the point $a = \dfrac 23$: for each $x \in [0,1]$ the Taylor formula states $$f(x) = f(\tfrac 23) + f'(\tfrac 23)(x-\tfrac 23) + \tfrac 12\int_{2/3}^x f''(t)(x-t)^2 \, dt.$$ Observe $$\int_0^1 x \, dx = \frac 12 \quad \text{and} \quad \int_0^1 x(x-\tfrac 23) \, dx = 0.$$ Thus $$\int_0^1 xf(x) \, dx = \tfrac 12 f(\tfrac 23) + \tfrac 12 \int_0^1 \int_{2/3}^x f''(t) x (x-t)^2 \, dt dx.$$

Can you apply a mean value theorem to the last integral?