Question Number 63710 by aliesam last updated on 07/Jul/19
Commented by mathmax by abdo last updated on 08/Jul/19
![let A_n =∫_0 ^(1/n) n^2 x^(x+1) ex ⇒A_n =∫_0 ^(1/n) n^2 e^((x+1)ln(x)) dx by parts u^′ =n^2 and v=e^((x+1)ln(x)) ⇒u=n^2 x and v^′ =(ln(x)+((x+1)/x))e^((x+1)ln(x)) ⇒A_n =[n^2 x x^(x+1) ]_0 ^(1/n) −∫_0 ^(1/n) n^2 x (ln(x)+1+(1/x))x^(x+1) dx =n((1/n))^((1/n)+1) −n^2 ∫_0 ^(1/n) (xlnx+x+1)x^(x+1) dx =(n/n^((1/n)+1) ) −n^2 ∫_0 ^(1/n) x^(x+2) ln(x) −n^2 ∫_0 ^(1/n) x^(x+2) dx−n^2 ∫_0 ^(1/n) x^(x+1) dx ⇒ 2A_n =(n/n^((1/n)+1) ) −n^2 ∫_0 ^(1/n) x^(x+2) ln(x)dx −n^2 ∫_0 ^(1/n) x^(x+2) dx we have lim_(n→+∞) (n/n^((1/n)+1) ) =1 rest to prove that lim_(n→+∞) n^2 {∫_0 ^(1/n) x^(x+2) ln(x)dx +∫_0 ^(1/n) x^(x+2) dx} =0 ⇒ lim_(n→+∞) A_n =(1/2) ...be continued...](Q63719.png)
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} {n}^{\mathrm{2}} \:{x}^{{x}+\mathrm{1}} \:{ex}\:\Rightarrow{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{n}^{\mathrm{2}} {e}^{\left({x}+\mathrm{1}\right){ln}\left({x}\right)} {dx}\:{by}\:{parts} \\ $$$${u}^{'} ={n}^{\mathrm{2}} \:{and}\:{v}={e}^{\left({x}+\mathrm{1}\right){ln}\left({x}\right)} \:\Rightarrow{u}={n}^{\mathrm{2}} {x}\:{and}\:{v}^{'} =\left({ln}\left({x}\right)+\frac{{x}+\mathrm{1}}{{x}}\right){e}^{\left({x}+\mathrm{1}\right){ln}\left({x}\right)} \\ $$$$\Rightarrow{A}_{{n}} =\left[{n}^{\mathrm{2}} {x}\:{x}^{{x}+\mathrm{1}} \right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{n}^{\mathrm{2}} {x}\:\left({ln}\left({x}\right)+\mathrm{1}+\frac{\mathrm{1}}{{x}}\right){x}^{{x}+\mathrm{1}} {dx} \\ $$$$={n}\left(\frac{\mathrm{1}}{{n}}\right)^{\frac{\mathrm{1}}{{n}}+\mathrm{1}} \:\:−{n}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:\left({xlnx}+{x}+\mathrm{1}\right){x}^{{x}+\mathrm{1}} {dx} \\ $$$$=\frac{{n}}{{n}^{\frac{\mathrm{1}}{{n}}+\mathrm{1}} }\:−{n}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{x}^{{x}+\mathrm{2}} {ln}\left({x}\right)\:−{n}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{x}^{{x}+\mathrm{2}} {dx}−{n}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{x}^{{x}+\mathrm{1}} {dx}\:\Rightarrow \\ $$$$\mathrm{2}{A}_{{n}} =\frac{{n}}{{n}^{\frac{\mathrm{1}}{{n}}+\mathrm{1}} }\:−{n}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{x}^{{x}+\mathrm{2}} {ln}\left({x}\right){dx}\:−{n}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{x}^{{x}+\mathrm{2}} {dx}\:{we}\:{have}\: \\ $$$${lim}_{{n}\rightarrow+\infty} \:\frac{{n}}{{n}^{\frac{\mathrm{1}}{{n}}+\mathrm{1}} }\:=\mathrm{1}\:\:{rest}\:{to}\:{prove}\:{that}\: \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{n}^{\mathrm{2}} \:\left\{\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{x}^{{x}+\mathrm{2}} {ln}\left({x}\right){dx}\:+\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:{x}^{{x}+\mathrm{2}} {dx}\right\}\:=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:...{be}\:{continued}... \\ $$