An other total-variation inequality of mine 
Given differentiable, continuous $f\left ( x \right )$ on the interval $\left [ 0, 1 \right ]$ so that $\int_{0}^{1}f\left ( x \right ){\rm d}x= 0.$ Prove that
$$\left | \int_{0}^{1}xf\left ( x \right ){\rm d}x \right |\leq \frac{1}{12}\max\left | {f}'\left ( x \right ) \right |$$

I think I should transform the constant $1/12$ into an integral like $k\int_{0}^{1}x^{2}{\rm d}x,$ but $k$ is very unusual, I need to your helps, even an example of $f\left ( x \right )$ so that $\int_{0}^{1}f\left ( x \right ){\rm d}x= 0$ in order to know what I must do with the constant. Thanks a real lot.
 A: $$\int_0^1 xf(x)\,dx=\int_0^1\left(x-\frac12\right)f(x)\,dx=\underbrace{-\frac12 x(1-x)f(x)\Bigg|_0^1}_{=0}+\frac12\int_0^1 x(1-x)f'(x)\,dx.$$ Hence, if $\displaystyle M=\sup_{x\in(0,1)}|f'(x)|$, then $\displaystyle\left|\int_0^1 xf(x)\,dx\right|\leqslant\frac{M}{2}\int_0^1 x(1-x)\,dx=\frac{M}{12}$.
A: Since $\int_0^1f(x)\mathrm{d}x=0$, by integration by parts,
\begin{align*}
\int_0^1xf(x)\mathrm{d}x&=\int_0^1(x+C)f(x)\mathrm{d}x\\
&=\Big(\frac{1}{2}x^2+Cx\Big)f(x)\bigg|_0^1-\int_0^1\Big(\frac{1}{2}x^2+Cx\Big)f'(x)\mathrm{d}x\\
&=\Big(\frac{1}{2}+C\Big)f(1)-\int_0^1\Big(\frac{1}{2}x^2+Cx\Big)f'(x)\mathrm{d}x
\end{align*}
Since $f(1)$ is unknown, so we let $C=-\frac{1}{2}$, we get
$$\int_0^1xf(x)\mathrm{d}x=-\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)f'(x)\mathrm{d}x$$
Therefore,
\begin{align*}
\left|\int_0^1xf(x)\mathrm{d}x\right|
&=\left|-\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)f'(x)\mathrm{d}x\right|\\
&=\left|-f'(\xi)\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)\mathrm{d}x\right|, \quad x\in(0,1)\\
&\leqslant\frac{1}{12}\max_{x\in[0,1]}|f'(x)|
\end{align*}
