Question Number 207878 by luciferit last updated on 29/May/24
Answered by Berbere last updated on 30/May/24
$${not}\:{well}\:{defind} \\ $$$$\left.{g}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {f}\left({t}^{\mathrm{4}} \right)+\mathrm{4}{t}^{\mathrm{4}} {f}'\left({t}\right)\right){dt}? \\ $$$$ \\ $$
Answered by Berbere last updated on 30/May/24
$$\int{f}\left({x}^{\mathrm{4}} \right){dx}+\int\mathrm{4}{x}^{\mathrm{4}} {f}'\left({x}^{\mathrm{4}} \right){dx} \\ $$$$\int{x}.\mathrm{4}{x}^{\mathrm{3}} {f}'\left({x}^{\mathrm{4}} \right){dx}={xf}\left({x}^{\mathrm{4}} \right)−\int{f}\left({x}^{\mathrm{4}} \right){dx} \\ $$$${g}\left({x}\right)=\int{f}\left({x}^{\mathrm{4}} \right){dx}+{xf}\left({x}^{\mathrm{4}} \right)−\int{f}\left({x}^{\mathrm{4}} \right){dx}={xf}\left({x}^{\mathrm{4}} \right)+{c} \\ $$$${g}\left(\mathrm{2}\right)=\mathrm{2}{f}\left(\mathrm{16}\right)+{c}=\mathrm{4} \\ $$$${c}=−\mathrm{4} \\ $$$${g}\left(−\mathrm{2}\right)=−\mathrm{12} \\ $$