Question Number 36413 by abdo.msup.com last updated on 01/Jun/18
$${let}\:\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \sqrt{\mathrm{3}+{x}}{dx} \\ $$$$\left.\mathrm{1}\right){calculate}\:{lim}_{{n}\rightarrow+\infty} {I}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:{n}\:{I}_{{n}} \\ $$
Commented by abdo.msup.com last updated on 04/Jun/18
![1) we have 0≤x≤1 ⇒ 3≤3+x≤4 ⇒ (√3) ≤(√(3+x))≤2 ⇒ ∫_0 ^1 (√3) x^n dx≤I_n ≤∫_0 ^1 2x^n dx ⇒ ((√3)/(n+1)) ≤ I_n ≤ (2/(n+1)) →_(n→+∞) 0 so lim_(n→+∞) I_n =0 2) by parts I_n =[ (1/(n+1))x^(n+1) (√(3+x))]_0 ^1 −∫_0 ^1 (x^(n+1) /(n+1)) (1/(2(√(3+x))))dx =(2/(n+1)) −(1/(2(n+1))) ∫_0 ^1 (x^(n+1) /(√(3+x)))dx⇒ nI_n = ((2n)/(n+1)) −(n/(2n+2)) ∫_0 ^1 (x^(n+1) /(√(3+x)))dx but (√3)≤(√(3+x))≤2 ⇒ (1/2) ≤ (1/(√(3+x)))≤(√3) ⇒ (x^(n+1) /2) ≤ (x^(n+1) /(√(3+x))) ≤ (√3) x^(n+1) ⇒ ∫_0 ^1 (x^(n+1) /2)dx≤ ∫_0 ^1 (x^(n+1) /(√(3+x)))dx≤ (√3)∫_0 ^1 x^(n+1) dx ⇒ (1/(2(n+2))) ≤ ∫_0 ^1 (x^(n+1) /(√(3+x)))dx≤ ((√3)/(n+1)) so lim_(n→+∞) ∫_0 ^1 (x^(n+1) /(√(3+x)))dx =0 ⇒ lim n I_n =2 (n→+∞)](Q36732.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:\Rightarrow\:\mathrm{3}\leqslant\mathrm{3}+{x}\leqslant\mathrm{4}\:\Rightarrow \\ $$$$\sqrt{\mathrm{3}}\:\leqslant\sqrt{\mathrm{3}+{x}}\leqslant\mathrm{2}\:\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{3}}\:{x}^{{n}} {dx}\leqslant{I}_{{n}} \leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{2}{x}^{{n}} {dx} \\ $$$$\Rightarrow\:\frac{\sqrt{\mathrm{3}}}{{n}+\mathrm{1}}\:\:\leqslant\:{I}_{{n}} \leqslant\:\frac{\mathrm{2}}{{n}+\mathrm{1}}\:\rightarrow_{{n}\rightarrow+\infty} \mathrm{0}\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{by}\:{parts}\:\:{I}_{{n}} \:=\left[\:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \sqrt{\mathrm{3}+{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}+{x}}}{dx} \\ $$$$=\frac{\mathrm{2}}{{n}+\mathrm{1}}\:\:−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}^{{n}+\mathrm{1}} }{\sqrt{\mathrm{3}+{x}}}{dx}\Rightarrow \\ $$$${nI}_{{n}} \:=\:\frac{\mathrm{2}{n}}{{n}+\mathrm{1}}\:−\frac{{n}}{\mathrm{2}{n}+\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{{n}+\mathrm{1}} }{\sqrt{\mathrm{3}+{x}}}{dx}\:{but} \\ $$$$\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{3}}\leqslant\sqrt{\mathrm{3}+{x}}\leqslant\mathrm{2}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}+{x}}}\leqslant\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{{x}^{{n}+\mathrm{1}} }{\mathrm{2}}\:\leqslant\:\frac{{x}^{{n}+\mathrm{1}} }{\sqrt{\mathrm{3}+{x}}}\:\leqslant\:\sqrt{\mathrm{3}}\:{x}^{{n}+\mathrm{1}} \:\Rightarrow\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{{n}+\mathrm{1}} }{\mathrm{2}}{dx}\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{{n}+\mathrm{1}} }{\sqrt{\mathrm{3}+{x}}}{dx}\leqslant\:\sqrt{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}+\mathrm{1}} {dx} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{2}\right)}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{n}+\mathrm{1}} }{\sqrt{\mathrm{3}+{x}}}{dx}\leqslant\:\frac{\sqrt{\mathrm{3}}}{{n}+\mathrm{1}}\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{{n}+\mathrm{1}} }{\sqrt{\mathrm{3}+{x}}}{dx}\:=\mathrm{0}\:\Rightarrow \\ $$$${lim}\:{n}\:{I}_{{n}} \:=\mathrm{2}\:\left({n}\rightarrow+\infty\right) \\ $$