Show $\frac{m}{12} \le \int_0^1 xf(x)dx \le \frac{M}{12}$ Given:


*

*$f'$ is continuous on $[0,1]$

*$\int_0^1 f(x)dx=0$

*$m \le f' \le M$, where $m$ is minimum and $M$ is maximum of $f$ on $[0,1]$


Show:
$$\frac{m}{12} \le \int_0^1 xf(x)dx \le \frac{M}{12}$$
Thoughts:
I believe that we want to somehow use the mean value theorem, along with part 3, in order to solve this. I also think that we want to use integration by parts, but I'm not necessarily sure how. I think we want a product of two functions that is equal to $f'(x)$ (so maybe $1$ and $f'(x)$?), but I'm not sure about that.
Any help would be appreciated - I'm happy to provide clarification!
 A: Since $\int_0^1 f(x)\; dx = 0$, we may write your integral as
$$ \int_0^1 (x - a) f(x)\; dx $$
for any constant $a$. Integrating by parts (differentiating the $f(x)$) should then give you
$$ \int_0^1 (x-a) f(x)\; dx = \left(\frac{1}{2} - a\right) f(1) - \int_0^1  \left(\frac{x^2}{2} - a x\right) f'(x)\; dx $$
We don't know about $f(1)$, so it's reasonable to try $a = 1/2$ to make that term go away: your integral is now
$$ \frac{1}{2} \int_0^1 (x - x^2) f'(x)\; dx $$
Note that $ x - x^2 \ge 0$ for $0 \le x \le 1$.  Thus if $m \le f'(x) \le M$ on that interval,
$$\frac{m}{12} = \frac{m}{2} \int_0^1 (x-x^2) \; dx \le \frac{1}{2} \int_0^1 (x-x^2) f'(x) \; dx \le \frac{M}{2} \int_0^1 (x-x^2)\; dx = \frac{M}{12}$$ 
A: Here is an alternative solution (credit goes to my professor) to the one Robert Israel posted:
Since $f'$ is continuous for $0\le x\le 1$, the Mean Value Theorem implies:
$$m \le \frac{f(x)-f(0)}{x} \le M$$


*

*When $x=1$,
$$m \le f(1)-f(0) \le M$$

*and for $x>0$,
$$mx \le f(1)-f(0) \le Mx$$

*it follows that
$$mx^2 \le f(1)-f(0) \le Mx^2$$

*Integrating over $[0,1]$ yields
$$\frac {m}{3} \le \int_0^1 xf(x)- \frac{f(0)}{2}\le \frac{M}{3}$$

*Using the fact that $m \le f' \le M$, we have
$$mx\le xf'(x)\le Mx$$

*Integrating (5) over $[0,1]$ gives us
$$\frac {m}{2} \le f(1)\le \frac{M}{2}$$

*Subtract (6) from (1) to obtain
$$\frac {m}{2} \le -f(0)\le \frac{M}{2}$$

*which can be written as
$$\frac {m}{4} \le \frac{-f(0)}{2}\le \frac{M}{4}$$

*Then (4) - (7) yields
$$\frac{m}{12} \le \int_0^1 xf(x)dx \le \frac{M}{12}$$

