# Integral inequality given the bounds for derivative

Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuously differentiable function such that $$\int_0^1 f(x) \, dx=0$$ and $m \leq f'(x) \leq M$ on $(0,1)$. Prove that $$\frac{m}{12} \leq \int_0^1 xf(x) \, dx \leq \frac{M}{12}.$$

This is just the last bonus question in our test yesterday. I wasn't able to answer of course. Though I did verify it by letting $\displaystyle f(x)= \left( x−\frac{1}{2} \right)^3$ for which I found $0\le f′(x)\le 3/4$ on $(0,1)$ and $$0 \le \int^1_0 xf(x) \, dx=\frac{1}{80} \le \frac{3}{48}.$$

I do not know how to prove this.

• So the integral is assumed to be $0$, but then you want to estimate that same integral?
– user147263
Jan 4, 2015 at 15:21
• This is just the last bonus question in our test yesterday. I wasn't able to answer of course. Though I did verify it by letting $f(x)=(x-1/2)^3$ for which I found $0\leq f'(x) \leq 3/4$ on (0,1) and $0\leq \int_0^1 xf(x)dx=\frac{1}{80} \leq \frac{3}{48}.$ I do not how to prove this. thanks Jan 4, 2015 at 15:41
• sorry for the typo... Jan 4, 2015 at 15:42

One proof goes via integration by parts. I show you how to start:

\begin{align} \int_0^1 1 \cdot x f(x)\,dx &= [(x-1)xf(x)]_0^1-\int_0^1(x-1)(f(x)+xf'(x))\,dx\\ &=-\int_0^1 xf(x)\, dx+\int_0^1(1-x)xf'(x)\, dx. \end{align} Here, we used that $\int_0^1f(x)\,dx=0$. Thus, we get $$2\int_0^1 xf(x)\,dx = \int_0^1 x(1-x)f'(x)\,dx.$$ Can you proceed from here?

Update:

Next, since $x(1-x)>0$ on $(0,1)$ and $m\leq f'(x)\leq M$, $$m\int_0^1 x(1-x)\,dx \leq \int_0^1 x(1-x)f'(x)\,dx\leq M\int_0^1 x(1-x)\,dx.$$ The argument is complete by noting that $\int_0^1 x(1-x)\,dx = \frac{1}{6}$.

(By the way, $f(x)=x-1/2$ gives the sharp estimate in both ends. Indeed, then $f'(x)=1$ so $m=M=1$. The integral $\int_0^1 xf(x)\,dx=1/12$.)

• How does the condition $m\leq f'(x) \leq M$ be useful in this? I did understand the integration by part though bu then I still have no idea. I have also tried that $f(x)=x-1/2$ during the test but I ended using a cubic function so I can have values for $m$ and $M$.. Jan 5, 2015 at 0:30
• I added the rest of it. Jan 5, 2015 at 6:50
• ok thank you very much mickep..i understood your solution completely... Jan 5, 2015 at 12:41