Integration Questions

Question Number 58937 by Tony Lin last updated on 02/May/19

$$\int_{\mathrm{0}} ^{\mathrm{2}} \underset{\frac{\mathrm{1}}{{n}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }{dx}=\:? \\$$

Commented by maxmathsup by imad last updated on 02/May/19

$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{2}} \:{lim}_{{n}\rightarrow+\infty} \:\:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx}\:\:{and}\:\:{f}_{{n}} \left({x}\right)\:=\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} } \\$$$$\left({f}_{{n}} \right){are}\:{continues}\:{on}\left[\mathrm{0},\mathrm{2}\right]\:\:\:{and}\:\mid{f}_{{n}} \left({x}\right)\mid\:\leqslant\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)\:{on}\:\left[\mathrm{0},\mathrm{2}\right]\:\Rightarrow \\$$$${A}_{{n}} ={lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:{f}_{{n}} \left({x}\right){dx}\:\:{but}\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:{f}_{{n}} \left({x}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }{dx} \\$$$$\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx}\:\:\:{we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\left(\mathrm{2}−{x}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}{x}−{x}^{\mathrm{2}} \right){dx}\:=\left[{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{2}}{\mathrm{3}} \\$$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx}\:=_{{x}\:=\mathrm{1}+{t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{2}−\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\:\:+\left(\mathrm{1}+{t}\right)^{{n}} \right)}{\mathrm{1}+\left(\mathrm{1}+{t}\right)^{{n}} }\:{dt} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\:+\left(\mathrm{1}+{t}\right)^{{n}} \right)}{\left(\mathrm{1}+{t}\right)^{{n}} \:+\mathrm{1}}\:{dt}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(...\right){dt} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\frac{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\:+\left(\mathrm{1}+{t}\right)^{{n}} \right)}{\left(\mathrm{1}+{t}\right)^{{n}} \:+\mathrm{1}}\:{dt} \\$$$$\frac{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\:+\left({t}+\mathrm{1}\right)^{{n}} \right)}{\left({t}+\mathrm{1}\right)^{{n}} \:+\mathrm{1}}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} \:+\left(\mathrm{1}−{t}\right)\left({t}+\mathrm{1}\right)^{{n}} }{\left({t}+\mathrm{1}\right)^{{n}} \:+\mathrm{1}}\:=\frac{\left({t}+\mathrm{1}\right)^{{n}} \left\{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)^{{n}} }\:+\mathrm{1}−{t}\right\}}{\left({t}+\mathrm{1}\right)^{{n}} \left\{\mathrm{1}+\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{{n}} }\right\}} \\$$$$=\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)^{{n}} }\:+\mathrm{1}−{t}}{\mathrm{1}\:+\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{{n}} }}\:\rightarrow\mathrm{1}−{t}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right){dt} \\$$$$=\left[{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{A}_{{n}} =\:\frac{\mathrm{2}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{7}}{\mathrm{6}}\:\: \\$$

Commented by Tony Lin last updated on 03/May/19

$${thanks},{sir} \\$$