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Let $f′(x)$ be continuous on $[0,1]$. Prove that

$$\int_0^1 |f(x)|dx \leq \max \left\{ \int_0^1 \lvert f′(x) \rvert dx,\bigg|\int_0^1 f(x)dx\bigg| \right\} $$

I've tried using the Intermediate Value Theorem and Fundamental Theorem of Calculus but the absolute values are really tripping me up.

When I use the Fundamental Theorem of Calculus, do I introduce a new $F(x)$ here such that $F'(x) = f(x)$?

Also, would I want to introduce another function $g(x) = |x|$ or $g(x) = |f(x)|$ and use $g(f(x))$ or $g(x)$. I'm struggling with finding the correct approach.

I think I need to try two cases: when $f$ changes signs on $[0,1]$ and when it does not. Any help would be really appreciated.

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1 Answer 1

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If $f(c)=0$ for some $c$ then $\int_0^{1}|f(x)|dx=\int_0^{1}|\int_c^{x} f'(t)dt|dx\leq \int_0^{1}\int_c^{x} |f'(t)|dtdx$. You can write this as $\int_c^{1} \int_t^{1} |f'(t)|dxdt=\int_c^{1} (1-t) |f'(t)|dt$ which does not exceed $\int_0^{1}|f'(t)|dt$.

If there is no such $c$ then $f$ is always positive or always negative so the result is obvious.

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