# Proving area under curve

Hey guys I've been stuck on this problem for a while and I've still got no headway into how I'm supposed to do this. Any tips or guiding points would be helpful.

Let $f$ be a differentiable function such that $f′$ is continuous on $[0,1]$, and $M$ the maximum value of $|f′(x)|$ on $[0,1]$. Prove that if $f(0) = f(1) = 0$, then

$$\int_0^1 |f(x)| dx\leq\frac{M}{4}$$

I've come to a conclusion via Rolle's theorem that there exists a c between 0 and 1 where f'(c) = 0 and that would be a local extrema. But nothing past that other than the fact that the max value of M is 1.

• Try to figure out which function will give an equality (Actually, there’s no &C^1$function s.t. equality holds.) – Seewoo Lee Apr 26 '18 at 10:36 • hmm is there a way to illustrate what you're saying? because I don't really understand – Jonathan Low Apr 26 '18 at 10:47 ## 1 Answer By the mean value theorem, for all$x\in (0,1)$, it is$\displaystyle{\frac{f(x)}{x}=\frac{f(x)-f(0)}{x-0}=f'(c_x)}$for some$c_x\in(0,x)$and similarly$\displaystyle{\frac{f(x)}{1-x}}=\frac{f(x)-f(1)}{1-x}=f'(d_x)$for some$d_x\in(x,1)$. Therefore, since$f(x)=xf'(c_x)$and$f(x)=(1-x)f'(d_x)$, by taking absolute values we get that for all$x\in(0,1)$, it is$|f(x)|\leq Mx$and$|f(x)|\leq M(1-x)$. Hence$|f(x)|\leq M\min\{x,1-x\}:=Mg(x)$.$g(x)$is a piecewise function, equal to$x$on$[0,\frac{1}{2}]$and equal to$1-x$on$[\frac{1}{2},1]\$.

By integrating the preceding inequality, one gets $$\int_0^1|f(x)|dx\leq M \bigg{(}\int_0^\frac{1}{2}xdx+\int_\frac{1}{2}^1(1-x)dx\bigg{)}=\frac{M}{4}$$

• Thanks so much! Took awhile to understand the last part on g(x) being a piecewise function but it's a clear now. Thanks! – Jonathan Low Apr 26 '18 at 14:11