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 75577 by Tony Lin last updated on 13/Dec/19

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xdxisdivergentorconvergent?$$

Answered by MJS last updated on 13/Dec/19

$$\underset{-\mathrm{\infty }}{\stackrel{+\mathrm{\infty }}{\int }}xdx=\underset{r\to \mathrm{\infty }}{\lim }\left(\underset{-r}{\stackrel{r}{\int }}xdx\right)=\underset{r\to \mathrm{\infty }}{\lim }\left({\left[\frac{x^2}{2}\right]}_{-r}^r\right)=\underset{r\to \mathrm{\infty }}{\lim 0}=0$$

Commented by MJS last updated on 13/Dec/19

$$\mathrm{no}$$

Commented by Tony Lin last updated on 13/Dec/19

$$thankssir,but$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}xdx={\int }_{-\mathrm{\infty }}^{0}xdx+{\int }_0^{+\mathrm{\infty }}xdx$$$$both{\int }_{-\mathrm{\infty }}^{0}xdx\mathrm{\&}{\int }_0^{+\mathrm{\infty }}xdxaredivergent$$$$Cantwodivergentintegralsadding$$$$togetherbetheconvergentintegral?$$

Commented by MJS last updated on 13/Dec/19

$$\mathrm{yes}.\mathrm{another}\mathrm{example}:$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x+1}=+\mathrm{\infty }$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x}=+\mathrm{\infty }$$$$\underset{x\to \mathrm{\infty }}{\lim }(\sqrt{x+1}-\sqrt{x})=0$$

Commented by vishalbhardwaj last updated on 13/Dec/19

$$\mathrm{sir}\mathrm{Is}\mathrm{this}\mathrm{possible}\mathrm{in}\mathrm{every}\mathrm{infinte}$$$$\mathrm{number}$$

Commented by vishalbhardwaj last updated on 13/Dec/19

$$\mathrm{sir}\mathrm{Is}\mathrm{this}\mathrm{possible}\mathrm{in}\mathrm{every}\mathrm{infinte}$$$$\mathrm{number}$$

Commented by Tony Lin last updated on 14/Dec/19

$$thankssir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com