How prove this inequality $\left|\int_{a}^{b}f(x)dx\right|\leq \frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx$ Let $f(x)\in \mathcal C^2([a,b]),f(\frac{a+b}{2})=0$.
Show that$$\left|\int_{a}^{b}f(x)dx\right|\leq \frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx.$$
I try to use  Taylor's Theorem with Integral form  of the Remainder,then I have $$f(x)=f(\frac{a+b}{2})+f'(\frac{a+b}{2})(x-\frac{a+b}{2})+\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt.$$ Since $f(\frac{a+b}{2})=\int_{a}^{b}f'(\frac{a+b}{2})(x-\frac{a+b}{2})dx=0, $ we get$\int_{a}^{b}f(x)dx=\int_{a}^{b}\left(\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt\right)dx.$
I think the point of this problem is to prove:
$$\left|\int_{a}^{b}\left(\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt\right)dx\right|\leq\frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx.$$The coeffient of $\frac{1}{8}$ is strange ,and how find it ? 
 A: Changing the order of integration, we obtain
\begin{align}
\int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx
&= \int_{\frac{a+b}{2}}^b \int_t^b (x-t)\,dx\, f''(t)\,dt\\
&= \int_{\frac{a+b}{2}}^b \frac{(b-t)^2}{2}f''(t)\,dt.
\end{align}
Now we can take the absolute value to get
\begin{align}
\biggl\lvert \int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert
&\leqslant \int_{\frac{a+b}{2}}^b \frac{(b-t)^2}{2} \lvert f''(t)\rvert\,dt\\
&\leqslant \frac{(b-a)^2}{8} \int_{\frac{a+b}{2}}^b \lvert f''(t)\rvert\,dt,
\end{align}
since $0 \leqslant b-t \leqslant b - \frac{a+b}{2} = \frac{b-a}{2}$ for $\frac{a+b}{2} \leqslant t \leqslant b$.
The analogous estimate for the integral over the first half of the interval gives the desired conclusion
\begin{align}
\biggl\lvert \int_{a}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx\biggr\rvert
&\leqslant \biggl\lvert \int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert
+ \biggl\lvert \int_a^{\frac{a+b}{2}} \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert\\
&\leqslant \frac{(b-a)^2}{8} \Biggl(\int_{\frac{a+b}{2}}^b \lvert f''(t)\rvert\,dt + \int_a^{\frac{a+b}{2}}\lvert f''(t)\rvert\,dt\Biggr).
\end{align}
