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 =∫_0 ^2 lim_(n→+∞) (((2−x)(x+x^n ))/(1+x^n )) dx and f_n (x) =(((2−x)(x+x^n ))/(1+x^n )) (f_n )are continues on[0,2] and ∣f_n (x)∣ ≤(2−x)(x+x^n ) on [0,2] ⇒ A_n =lim_(n→+∞) ∫_0 ^2 f_n (x)dx but ∫_0 ^2 f_n (x)dx =∫_0 ^1 (((2−x)(x+x^n ))/(1+x^n ))dx +∫_1 ^2 (((2−x)(x+x^n ))/(1+x^n )) dx we have lim_(n→+∞) ∫_0 ^1 (((2−x)(x+x^n ))/(1+x^n )) dx =∫_0 ^1 x(2−x)dx =∫_0 ^1 (2x−x^2 )dx =[x^2 −(x^3 /3)]_0 ^1 =1−(1/3) =(2/3) ∫_1 ^2 (((2−x)(x+x^n ))/(1+x^n )) dx =_(x =1+t) ∫_0 ^1 (((2−1−t)(1+t +(1+t)^n ))/(1+(1+t)^n )) dt =∫_0 ^1 (((1−t)(1+t +(1+t)^n ))/((1+t)^n +1)) dt ⇒lim_(n→+∞) ∫_0 ^1 (...)dt =∫_0 ^1 lim_(n→+∞) (((1−t)(1+t +(1+t)^n ))/((1+t)^n +1)) dt (((1−t)(1+t +(t+1)^n ))/((t+1)^n +1)) =((1−t^2 +(1−t)(t+1)^n )/((t+1)^n +1)) =(((t+1)^n {((1−t^2 )/((t+1)^n )) +1−t})/((t+1)^n {1+(1/((t+1)^n ))})) =((((1−t^2 )/((t+1)^n )) +1−t)/(1 +(1/((t+1)^n )))) →1−t ⇒lim_(n→+∞) ∫_1 ^2 (((2−x)(x+x^n ))/(1+x^n )) dx =∫_0 ^1 (1−t)dt =[t−(t^2 /2)]_0 ^1 =1−(1/2) =(1/2) ⇒ A_n = (2/3) +(1/2) =(7/6)](https://www.tinkutara.com/question/Q58998.png)
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{2}} \:{lim}_{{n}\rightarrow+\infty} \:\:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx}\:\:{and}\:\:{f}_{{n}} \left({x}\right)\:=\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} } \\ $$$$\left({f}_{{n}} \right){are}\:{continues}\:{on}\left[\mathrm{0},\mathrm{2}\right]\:\:\:{and}\:\mid{f}_{{n}} \left({x}\right)\mid\:\leqslant\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)\:{on}\:\left[\mathrm{0},\mathrm{2}\right]\:\Rightarrow \\ $$$${A}_{{n}} ={lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:{f}_{{n}} \left({x}\right){dx}\:\:{but}\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:{f}_{{n}} \left({x}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }{dx} \\ $$$$\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx}\:\:\:{we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\left(\mathrm{2}−{x}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}{x}−{x}^{\mathrm{2}} \right){dx}\:=\left[{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx}\:=_{{x}\:=\mathrm{1}+{t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{2}−\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\:\:+\left(\mathrm{1}+{t}\right)^{{n}} \right)}{\mathrm{1}+\left(\mathrm{1}+{t}\right)^{{n}} }\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\:+\left(\mathrm{1}+{t}\right)^{{n}} \right)}{\left(\mathrm{1}+{t}\right)^{{n}} \:+\mathrm{1}}\:{dt}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(…\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\frac{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\:+\left(\mathrm{1}+{t}\right)^{{n}} \right)}{\left(\mathrm{1}+{t}\right)^{{n}} \:+\mathrm{1}}\:{dt} \\ $$$$\frac{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\:+\left({t}+\mathrm{1}\right)^{{n}} \right)}{\left({t}+\mathrm{1}\right)^{{n}} \:+\mathrm{1}}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} \:+\left(\mathrm{1}−{t}\right)\left({t}+\mathrm{1}\right)^{{n}} }{\left({t}+\mathrm{1}\right)^{{n}} \:+\mathrm{1}}\:=\frac{\left({t}+\mathrm{1}\right)^{{n}} \left\{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)^{{n}} }\:+\mathrm{1}−{t}\right\}}{\left({t}+\mathrm{1}\right)^{{n}} \left\{\mathrm{1}+\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{{n}} }\right\}} \\ $$$$=\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)^{{n}} }\:+\mathrm{1}−{t}}{\mathrm{1}\:+\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{{n}} }}\:\rightarrow\mathrm{1}−{t}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\left(\mathrm{2}−{x}\right)\left({x}+{x}^{{n}} \right)}{\mathrm{1}+{x}^{{n}} }\:{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right){dt} \\ $$$$=\left[{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{A}_{{n}} =\:\frac{\mathrm{2}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{7}}{\mathrm{6}}\:\: \\ $$
Commented by Tony Lin last updated on 03/May/19
$${thanks},{sir} \\ $$