Previous in Relation and Functions Next in Relation and Functions | ||
Question Number 57416 by Abdo msup. last updated on 03/Apr/19 | ||
$${let}\:{f}\left({x}\right)=\int_{\mathrm{2}{x}} ^{\mathrm{4}{x}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{3}} \\ $$$$\left.\mathrm{1}\right){find}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)\:{and}\:{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right) \\ $$ | ||
Commented by maxmathsup by imad last updated on 04/Apr/19 | ||
$$\left.\mathrm{1}\right){we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{2}{x}} ^{\mathrm{4}{x}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}+\mathrm{2}}\:=\int_{\mathrm{2}{x}} ^{\mathrm{4}{x}} \:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{2}}\:=_{{t}−\mathrm{1}=\sqrt{\mathrm{2}}{u}} \:\:\int_{\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\mathrm{4}{x}−\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\:\:\frac{\sqrt{\mathrm{2}}{du}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\left[{arctanu}\right]_{\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\frac{\mathrm{4}{x}−\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\:\:\:\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left\{{arctan}\left(\frac{\mathrm{4}{x}−\mathrm{1}}{\sqrt{\mathrm{2}}}\right)−{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right\} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left\{\:−{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\:+{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right\}=\mathrm{0}\:\:{also} \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left\{\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{2}}\right\}\:=\mathrm{0}\:. \\ $$ | ||