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Question Number 78635 by Pratah last updated on 19/Jan/20

Commented by john santu last updated on 19/Jan/20

$$\underset{n\to \mathrm{\infty }}{\lim }\left(\frac{\underset{n}{\stackrel{2n}{\int }}\left(\frac{x}{x^5+1}\right)dx}{n^{-3}}\right)=\underset{n\to \mathrm{\infty }}{\lim }\frac{\left(\frac{2n}{32n^5+1}\right).2-\left(\frac{n}{n^5+1}\right).1}{-3n^{-4}}$$$$=-\frac{1}{3}\underset{n\to \mathrm{\infty }}{\lim }\left(\frac{4n^5}{32n^5+1}\right)-\left(\frac{n^5}{n^5+1}\right)$$$$=-\frac{1}{3}(\frac{1}{8}-1)=-\frac{1}{3}\times \left(\frac{-7}{8}\right)=\frac{7}{24}.$$

Commented by Pratah last updated on 19/Jan/20

$$\mathrm{thanks}$$

Commented by mr W last updated on 19/Jan/20

$$itistoproveatfirstthat\underset{n\to \mathrm{\infty }}{\lim }\underset{n}{\stackrel{2n}{\int }}\left(\frac{x}{x^5+1}\right)dx=0!$$

Commented by jagoll last updated on 19/Jan/20

$${\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{can}\mathrm{be}\mathrm{using}$$$$\mathrm{for}\mathrm{limit}\mathrm{form}\frac{\mathrm{\infty }}{\mathrm{\infty }}$$

Commented by mr W last updated on 19/Jan/20

$$wehavehere\frac{0}{0}$$

Commented by john santu last updated on 19/Jan/20

$$inotherway$$$$letn=\frac{1}{t}\Rightarrow t\to 0$$$$\underset{t\to 0}{\lim }\left(\frac{{\underset{\frac{1}{t}}{\int }}^{\frac{2}{t}}\left(\frac{x}{x^5+1}\right)dx}{t^3}\right)=$$$$\underset{t\to 0}{\lim }[\frac{\left(\frac{2}{t}\right).\left(\frac{-2}{t^2}\right)}{(\frac{32}{t^5}+1)}-\frac{\left(\frac{1}{t}\right).\left(\frac{-1}{t^2}\right)}{(\frac{1}{t^5}+1)}]\times \frac{1}{3t^2}$$$$=[\frac{-4t^2}{32+t^5}-\frac{(-t^2)}{(1+t^5)}]\times \frac{1}{3t^2}$$$$=[\frac{-1}{8}+1]\times \frac{1}{3}=\frac{7}{24}.★$$

Commented by mr W last updated on 19/Jan/20

$$I_n=\underset{n}{\stackrel{2n}{\int }}\left(\frac{x}{x^5+1}\right)dx$$$$0<\frac{x}{x^5+1}<\frac{x}{x^5}=\frac{1}{x^4}$$$$0<I_n<\frac{1}{{\left(2n\right)}^4}=\frac{1}{32n^4}$$$$0<\underset{n\to \mathrm{\infty }}{\lim }{I}_n<\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{32n^4}=0$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }{I}_n=0$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\frac{\underset{n}{\stackrel{2n}{\int }}\left(\frac{x}{x^5+1}\right)dx}{\frac{1}{n^3}}=\frac{0}{0}$$

Commented by john santu last updated on 19/Jan/20

$$byusingthesequencetheoremsir$$

Answered by mr W last updated on 19/Jan/20

$$I={\int }_n^{2n}\frac{x}{1+x^5}dx$$$$lett=\frac{1}{x}$$$$dx=-\frac{dt}{t^2}$$$$I={\int }_{\frac{1}{n}}^{\frac{1}{2n}}\frac{\frac{1}{t}}{1+\frac{1}{t^5}}\times (-\frac{1}{t^2})dt$$$$={\int }_{\frac{1}{2n}}^{\frac{1}{n}}\frac{t^2}{1+t^5}dt$$$$\underset{n\to \mathrm{\infty }}{\lim }{n}^3{\int }_n^{2n}\frac{x}{1+x^5}dx$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{{\int }_n^{2n}\frac{x}{1+x^5}dx}{{\left(\frac{1}{n}\right)}^3}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{{\int }_{\frac{1}{2n}}^{\frac{1}{n}}\frac{t^2}{1+t^5}dt}{{\left(\frac{1}{n}\right)}^3}$$$$=\underset{u\to 0}{\lim }\frac{{\int }_{\frac{u}{2}}^u\frac{t^2}{1+t^5}dt}{u^3}(=\frac{0}{0})$$$$=\underset{u\to 0}{\lim }\frac{\frac{u^2}{1+u^5}-\frac{{\left(\frac{u}{2}\right)}^2}{1+{\left(\frac{u}{2}\right)}^5}\times \frac{1}{2}}{3u^2}$$$$=\underset{u\to 0}{\lim }\frac{(\frac{1}{1+u^5}-\frac{4}{32+u^5})}{3}$$$$=\frac{7}{8\times 3}$$$$=\frac{7}{24}$$

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