Question Number 130255 by stelor last updated on 23/Jan/21
$$\mathrm{please}\:\:….\:\mathrm{I}\:\mathrm{am}\:\mathrm{confiouse}….\:\:\:\mathrm{1}\:\mathrm{and}\:\mathrm{2}. \\ $$$$\mathrm{1}.\:\:\:\:\:\:\:\int\mid{x}\mid\mathrm{d}{x}\:=\:?\:? \\ $$$$\mathrm{2}.\:\:\:\:\:\mathrm{with}\:\left(\mathrm{F}\left({x}\right)\right)^{'} \:=\mathrm{f}\left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\mid\mathrm{f}\left({x}\right)\mid{dx}\:=\:?? \\ $$$$ \\ $$
Answered by Olaf last updated on 23/Jan/21
$$\int\mid{x}\mid{dx}\:=\:\begin{cases}{−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{C},\:{x}\:\leqslant\:\mathrm{0}}\\{+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{C},\:{x}\:\geqslant\:\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:\int\mid{x}\mid{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\mid{x}\mid+\mathrm{C} \\ $$$$ \\ $$$$\int\mid{f}\left({x}\right)\mid{dx}\:=\:\begin{cases}{−\mathrm{F}\left({x}\right)+\mathrm{C},\:{f}\left({x}\right)\:\leqslant\:\mathrm{0}}\\{+\mathrm{F}\left({x}\right)+\mathrm{C},\:{f}\left({x}\right)\:\geqslant\:\mathrm{0}}\end{cases} \\ $$