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Question Number 445 by 123456 last updated on 05/Jan/15

0≤∣f(x)∣≤∣g(x)∣  ∫g(x)−f(x) dx≤∫f(x)+g(x) dx  ?

$$\mathrm{0}\leqslant\mid{f}\left({x}\right)\mid\leqslant\mid{g}\left({x}\right)\mid \\ $$$$\int{g}\left({x}\right)−{f}\left({x}\right)\:{dx}\leqslant\int{f}\left({x}\right)+{g}\left({x}\right)\:{dx}\:\:? \\ $$

Commented by prakash jain last updated on 05/Jan/15

f(x)=−1  g(x)=1  ∣f(x)∣=1  ∣g(x)∣=1  g(x)−f(x)=2  f(x)+g(x)=0

$${f}\left({x}\right)=−\mathrm{1} \\ $$$${g}\left({x}\right)=\mathrm{1} \\ $$$$\mid{f}\left({x}\right)\mid=\mathrm{1} \\ $$$$\mid{g}\left({x}\right)\mid=\mathrm{1} \\ $$$${g}\left({x}\right)−{f}\left({x}\right)=\mathrm{2} \\ $$$${f}\left({x}\right)+{g}\left({x}\right)=\mathrm{0} \\ $$

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