Question Number 78717 by jagoll last updated on 20/Jan/20
$$\mathrm{given}\: \\ $$$$\int\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}\:\sqrt[{\mathrm{3}\:}]{\mathrm{g}\left(\mathrm{x}\right)}}\:.\: \\ $$$$\mathrm{g}'\left(\mathrm{1}\right)=\:\mathrm{g}\left(\mathrm{1}\right)\:=\:\mathrm{8}\:\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=? \\ $$$$ \\ $$
Commented by john santu last updated on 20/Jan/20
$$\int{f}\left({x}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({g}\left({x}\right)\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${f}\left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{6}}{g}'\left({x}\right)×\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}\:}]{\left({g}\left({x}\right)\right)^{\mathrm{4}} }} \\ $$$${f}\left(\mathrm{1}\right)=\:−\frac{\mathrm{1}}{\mathrm{6}}.\left(\mathrm{8}\right).\frac{\mathrm{1}}{\mathrm{16}}=−\frac{\mathrm{1}}{\mathrm{12}}\:\bigstar \\ $$
Commented by jagoll last updated on 20/Jan/20
$$\mathrm{thanks} \\ $$