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Question Number 111313 by bemath last updated on 03/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\left(1\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{{(\mathrm{n}+1)}^5+{(\mathrm{n}+2)}^5+{(\mathrm{n}+3)}^5+...+{\left(2\mathrm{n}\right)}^5}{{\mathrm{n}}^6}?$$$$\left(2\right)\int \frac{\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}{{\mathrm{x}}^4}\mathrm{dx}$$

Answered by ajfour last updated on 03/Sep/20

$$\left(2\right)x=a\sec \theta \Rightarrow \cos \theta =\frac{a}{x}$$$$\int \frac{a\tan \theta \left(a\sec \theta \tan \theta \right)d\theta }{a^4\sec {}^4\theta }$$$$=\frac{1}{a^2}\int \sin {}^2\theta \cos \theta d\theta =\frac{1}{a^2}\left(\frac{\sin {}^3\theta }{3}\right)+c$$$$=\frac{1}{3a^2}{(1-\frac{a^2}{x^2})}^{3/2}+c$$

Commented by bemath last updated on 03/Sep/20

$$\mathrm{thank}\mathrm{you}$$

Answered by bobhans last updated on 03/Sep/20

$$\left(◼\right)\mathrm{I}=\int \frac{\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}{{\mathrm{x}}^4}\mathrm{dx}$$$$[\mathrm{let}\mathrm{x}=\mathrm{a}\sec \mathrm{t}\Rightarrow \mathrm{dx}=\mathrm{asec}\mathrm{t}\tan \mathrm{t}\mathrm{dt}]$$$$\mathrm{I}=\int \frac{\sqrt{{\mathrm{a}}^2(\sec {}^2\mathrm{t}-1)}}{{\mathrm{a}}^4\sec {}^4\mathrm{t}}.\mathrm{a}\sec \mathrm{t}\tan \mathrm{t}\mathrm{dt}$$$$\mathrm{I}=\frac{1}{{\mathrm{a}}^2}\int \frac{{\tan }^2\mathrm{t}}{\sec {}^3\mathrm{t}}\mathrm{dt}=\frac{1}{{\mathrm{a}}^2}\int \sin {}^2\mathrm{t}\cos \mathrm{t}\mathrm{dt}$$$$\mathrm{I}=\frac{1}{{\mathrm{a}}^2}\int \sin {}^2\mathrm{t}\mathrm{d}\left(\sin \mathrm{t}\right)=\frac{1}{{3\mathrm{a}}^2}\sin {}^3\mathrm{t}+\mathrm{c}$$$$\mathrm{I}=\frac{1}{{3\mathrm{a}}^2}{\left(\frac{\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}{\mathrm{x}}\right)}^3+\mathrm{c}=\frac{1}{{3\mathrm{a}}^2{\mathrm{x}}^3}\sqrt{{({\mathrm{x}}^2-{\mathrm{a}}^2)}^3}+\mathrm{c}$$

Answered by bobhans last updated on 03/Sep/20

$$\left(★\right)\mathrm{L}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[\frac{1}{{\mathrm{n}}^6}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{(\mathrm{n}+\mathrm{k})}^5\right]$$$$\mathrm{L}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{(1+\frac{\mathrm{k}}{\mathrm{n}})}^5.\frac{1}{\mathrm{n}}]$$$$\mathrm{recall}\underset{\mathrm{a}}{\stackrel{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\mathrm{f}(\mathrm{x}+\mathrm{k}.△\mathrm{x}).△\mathrm{x}$$$$\mathrm{where}△\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}},\mathrm{we}\mathrm{have}△\mathrm{x}=\frac{1}{\mathrm{n}}$$$$\mathrm{by}\mathrm{letting}\to \{\begin{array}{l}\mathrm{a}=1\\ \mathrm{b}=2\end{array}$$$$\mathrm{therefore}\mathrm{L}=\underset{1}{\stackrel{2}{\int }}{\mathrm{x}}^5\mathrm{dx}={\left[\frac{1}{6}{\mathrm{x}}^6\right]}_1^2=\frac{63}{6}=\frac{21}{2}$$

Commented by bemath last updated on 03/Sep/20

$$\mathrm{yes}...\mathrm{thank}\mathrm{you}$$

Answered by Dwaipayan Shikari last updated on 03/Sep/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}({(1+\frac{1}{n})}^5+{(1+\frac{2}{n})}^5+....n)$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{r=1}{\sum ^n}{(1+\frac{r}{n})}^5$$$$={\int }_0^1{(1+x)}^{5}dx=\frac{1}{6}(64-1)=\frac{21}{2}$$

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