Prove $\left|\int_a^bf(x)dx \right|\le \frac{(b-a)^3}{24} \max_{a\le x\le b}|f''(x)|$ Suppose that $f\in C^2 [a, b]$ and that $f(\frac{a+b}{2}) =0$ then prove that \begin{equation}  \bigg|\int_a^bf(x)dx \bigg|\le \frac{(b-a)^3}{24} \max_{a\le x\le b}|f''(x)| \end{equation}.
I know that since $f\in C^2$, by the Taylor expansion of $f$, in a neighborhood of $x_0$ up to the 2nd derivative I can express $f$ as
$$f(x) = f(x_0) +\frac{(x-x_0)}{1!}f'(x_0)+\frac{(x-x_0)^2}{2!}f''(x_0)$$
I'm not sure if this is right, but  I go ahead to write the integral as
$$\bigg| \int_a^b f(x) dx\bigg|\le \max_{a\le x\le b}|f''(x) | \bigg| \int_a^b \frac{(x-x_0)^2} {2!} dx\bigg|. $$
But I don't seem to be getting what is needed because I have to use the condition $f(\frac{a+b} {2})$. What is wrong here with my approach?

Solution:
The Taylor expansion (with Lagrange remainder) of $f(x)$ at the point of $x=\dfrac{a+b}{2}$ is
\begin{align}
f(x)&=\color{blue}{\underbrace{f\left(\frac{a+b}{2}\right)}_{0}}+f^\prime\left(\frac{a+b}{2}\right)\left(x-\frac{a+b}{2}\right)+\frac{1}{2}f^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2\\
&=f^\prime\left(\frac{a+b}{2}\right)\left(x-\frac{a+b}{2}\right)+\frac{1}{2}f^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2,~~x\in[a,b],
\end{align}
where $\xi\in[a,b]$.
Integrate both sides, then we have
\begin{align}
\int_a^bf(x)dx&=f^\prime\left(\frac{a+b}{2}\right)\color{blue}{\underbrace{\int_a^b\left(x-\frac{a+b}{2}\right)dx}_{0}}+\frac{1}{2}\int_a^bf^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2dx\\
&=\frac{1}{2}\int_a^bf^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2dx.
\end{align}
Finally, there is
\begin{align}
\left|\int_a^bf(x)dx\right|&\le \frac{1}{2}\int_a^b\left|f^{\prime\prime}\left(\xi\right)\left(x-\frac{a+b}{2}\right)^2\right|dx\\
&\le\frac{A}{2}\int_a^b\left(x-\frac{a+b}{2}\right)^2dx=\frac{(b-a)^3}{24}A,
\end{align}
where $A=\underset{a\le x\le b}{\max}\left|f^{\prime\prime}(x)\right|$.
