Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 36413 by abdo.msup.com last updated on 01/Jun/18

$$let{I}_n={\int }_0^1{x}^n\sqrt{3+x}dx$$$$1)calculate{lim}_{n\to +\mathrm{\infty }}{I}_n$$$$2)calculate{lim}_{n\to +\mathrm{\infty }}n{I}_n$$

Commented by abdo.msup.com last updated on 04/Jun/18

$$1)wehave0⩽x⩽1\Rightarrow 3⩽3+x⩽4\Rightarrow$$$$\sqrt{3}⩽\sqrt{3+x}⩽2\Rightarrow {\int }_0^1\sqrt{3}{x}^{n}dx⩽I_n⩽{\int }_0^{1}2x^{n}dx$$$$\Rightarrow \frac{\sqrt{3}}{n+1}⩽I_n⩽\frac{2}{n+1}{\to }_{n\to +\mathrm{\infty }}0so$$$${lim}_{n\to +\mathrm{\infty }}{I}_n=0$$$$2)byparts{I}_n={\left[\frac{1}{n+1}{x}^{n+1}\sqrt{3+x}\right]}_0^1$$$$-{\int }_0^1\frac{x^{n+1}}{n+1}\frac{1}{2\sqrt{3+x}}dx$$$$=\frac{2}{n+1}-\frac{1}{2(n+1)}{\int }_0^1\frac{x^{n+1}}{\sqrt{3+x}}dx\Rightarrow$$$${nI}_n=\frac{2n}{n+1}-\frac{n}{2n+2}{\int }_0^1\frac{x^{n+1}}{\sqrt{3+x}}dxbut$$$$\sqrt{3}⩽\sqrt{3+x}⩽2\Rightarrow \frac{1}{2}⩽\frac{1}{\sqrt{3+x}}⩽\sqrt{3}$$$$\Rightarrow \frac{x^{n+1}}{2}⩽\frac{x^{n+1}}{\sqrt{3+x}}⩽\sqrt{3}{x}^{n+1}\Rightarrow$$$${\int }_0^1\frac{x^{n+1}}{2}dx⩽{\int }_0^1\frac{x^{n+1}}{\sqrt{3+x}}dx⩽\sqrt{3}{\int }_0^1{x}^{n+1}dx$$$$\Rightarrow \frac{1}{2(n+2)}⩽{\int }_0^1\frac{x^{n+1}}{\sqrt{3+x}}dx⩽\frac{\sqrt{3}}{n+1}so$$$${lim}_{n\to +\mathrm{\infty }}{\int }_0^1\frac{x^{n+1}}{\sqrt{3+x}}dx=0\Rightarrow$$$$limn{I}_n=2(n\to +\mathrm{\infty })$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com