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
![1)we have f(x)=∫_(2x) ^(4x) (dt/(t^2 −2t+1+2)) =∫_(2x) ^(4x) (dt/((t−1)^2 +2)) =_(t−1=(√2)u) ∫_((2x−1)/( (√2))) ^((4x−1)/( (√2))) (((√2)du)/(2(1+u^2 ))) =(1/( (√2))) [arctanu]_((2x−1)/( (√2))) ^((4x−1)/( (√2))) =(1/( (√2))){arctan(((4x−1)/( (√2))))−arctan(((2x−1)/( (√2))))} 2) we have lim_(x→0) f(x)=(1/( (√2))){ −arctan((1/( (√2)))) +arctan((1/( (√2))))}=0 also lim_(x→+∞) f(x) =(1/( (√2))){(π/2) −(π/2)} =0 .](https://www.tinkutara.com/question/Q57446.png)
$$\left.\mathrm{1}\right){we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{2}{x}} ^{\mathrm{4}{x}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}+\mathrm{2}}\:=\int_{\mathrm{2}{x}} ^{\mathrm{4}{x}} \:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{2}}\:=_{{t}−\mathrm{1}=\sqrt{\mathrm{2}}{u}} \:\:\int_{\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\mathrm{4}{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\:\frac{\sqrt{\mathrm{2}}{du}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[{arctanu}\right]_{\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\mathrm{4}{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{{arctan}\left(\frac{\mathrm{4}{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right\} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\:−{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:+{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right\}=\mathrm{0}\:\:{also} \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{2}}\right\}\:=\mathrm{0}\:. \\ $$