<s>[INST] Show that the inequality $\left|\int_{0}^{1} f(x)\,dx\right| \leq \frac{1}{12}$ holds for certain initial conditions 
Given that a function $f$ has a continuous second derivative on the interval $[0,1]$,  $f(0)=f(1)=0$, and $|f''(x)|\leq 1$, show that $$\left|\int_{0}^{1}f(x)\,dx\right|\leq \frac{1}{12}\,.$$

My attempt: This looks to be a maximization/minimization problem. Since the largest value $f''(x)$ can take on is $1$, then the first case will be to assume $f''(x)=1$. This is because it is the maximum concavity and covers the most amount of area from $[0,1]$ while still maintaining the given conditions.
Edit: Because of the MVT and Rolle's Theorem, there exists extrema on the interval $[0,1]$ satisfying $f'(c)=0$ for some $c\in[0,1]$. These extrema could occur at endpoints.
Then $f'(x)=x+b$ and $f(x)=\frac{x^2}{2}+bx+c$. Since $f(0)=0$, then $c=0$ and $f(1)=0$, then $b=-\frac{1}{2}$. Remark: Any function with a continuous, constant second derivative will be of the form $ax^2+bx+c$ and in this case, $a=-b$ and $c=0$. Now,  $$\begin{align*}\int_{0}^{1}f(x)\,dx&=\frac{1}{2}\int_{0}^{1}(x^2-x)\,dx\\&=\frac{1}{2}\bigg[\frac{x^3}{3}-\frac{x^2}{2}\bigg]_{x=0}^{x=1}\\&=-\frac{1}{12}\end{align*}$$
Next, we assume that $f''(x)=-1$ and repeating the process yields $$ \begin{align*}\int_{0}^{1}f(x)\,dx&=\frac{1}{2}\int_{0}^{1}(-x^2+x)\,dx\\&=\frac{1}{2}\bigg[\frac{-x^3}{3}+\frac{x^2}{2}\bigg]_{x=0}^{x=1}\\&=\frac{1}{12}\end{align*}$$ Thus we have shown that at the upper and lower bounds for $f''(x)$ that $\frac{-1}{12}\leq\int_{0}^{1}f(x)\,dx\leq \frac{1}{12} 
 \Longleftrightarrow \left|\int_{0}^{1}f(x)\,dx\right|\leq\frac{1}{12}$ because $f''(x)$ is continuous on $[0,1]$.
I was wondering if this was 'rigorous' enough to be considered a full proof and solution to the problem.
 [/INST] Consider the following integral:
$$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx. $$
By integrating by parts twice, you get
$$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx = \underbrace{\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f'(x)\bigg|_0^1}_{0} - \int_0^1\bigg(x-\frac{1}{2}\bigg)f'(x)dx=$$$$= - \int_0^1\bigg(x-\frac{1}{2}\bigg)f'(x)dx= \underbrace{- \bigg(x-\frac{1}{2}\bigg)f(x)\bigg|_0^1}_{0} + \int_0^1f(x)dx$$
Therefore,
$$\boxed{\int_{0}^{1}f(x)\, dx = \int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx}$$
Now use the following inequality:
$$\left|\int_{a}^{b}f(x)g(x)\,dx\right| \leq \int_{a}^{b}|f(x)||g(x)|\, dx$$
Since $g(x)=\frac{x^{2}}{2}-\frac{x}{2}$
is the expression you got, this should yield the desired result.
$$\left|\int_0^1 f(x)\,dx\right|=\left| \int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx\right|\le\frac{1}{2}\int_{0}^{1}|x^2-x|\,dx=\frac{1}{12}$$
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