# Difference between Absolute value of integral and integral of absolute value

I have this problem which I am not completely sure how to start it. It is the Fundamental Theorem of Calculus section of the textbook I am trying to teach myself from.

Let $f:[0,1]\rightarrow \mathbb{R}$ be differentiable on $(0,1)$. Suppose both $f$ and $f'$ are continuous on $[0,1]$. Prove that

$$\int_{0}^{1} |f(x)| dx - \Bigg|\int_{0}^{1} f(x) dx\Bigg|\leq \frac{1}{2} \max\limits_{x\in [0,1]}|f'(x)|.$$

This feels like it could be a use for the MVT(for integrals) which is also in this section. It says, for our case:

If $f$ is continuous on [0,1], then there exists a number $c$ such that $$f(c)=\int_{0}^{1} f(x) dx.$$

This is probably very easy, I am just not seeing it! Thanks for the help!

• Hint: from the mvt you have: $|f(c)|-|f(b)|\leq (b-c)|f'(d)|$ for some b,c,d. – YoTengoUnLCD Apr 14 '16 at 3:56

Hint: Let $M = \sup_{[0,1]} |f'|.$ By the mean value property of integrals of continuous functions, it suffices to show
$$\int_0^1|f(x)|\, dx -|f(c)| \le \frac{M}{2}$$
for any $c \in [0,1].$ The left side equals $\int_0^1(|f(x)| -|f(c)|)\, dx,$ which is $\le \int_0^1(|f(x) -f(c)|)\, dx.$