bemath-1-lim-n-n-1-5-n-2-5-n-3-5-2n-5-n-6-2-x-2-a-2-x-4-dx-

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

\sqrt{bemath}

(1) \lim_{n \to \infty} \frac{{(n + 1)}^{5} + {(n + 2)}^{5} + {(n + 3)}^{5} + \dots + {(2 n)}^{5}}{n^{6}} ?

(2) \int \frac{\sqrt{x^{2} - a^{2}}}{x^{4}} dx

Answered by ajfour last updated on 03/Sep/20

(2) x = a \sec θ \Rightarrow \cos θ = \frac{a}{x}

\int \frac{a \tan θ (a \sec θ \tan θ) d θ}{a^{4} \sec^{4} θ}

= \frac{1}{a^{2}} \int \sin^{2} θ \cos θ d θ = \frac{1}{a^{2}} (\frac{\sin^{3} θ}{3}) + c

= \frac{1}{3 a^{2}} {(1 - \frac{a^{2}}{x^{2}})}^{3 / 2} + c

Commented by bemath last updated on 03/Sep/20

thank you

Answered by bobhans last updated on 03/Sep/20

() I = \int \frac{\sqrt{x^{2} - a^{2}}}{x^{4}} dx

[let x = a \sec t \Rightarrow dx = asec t \tan t dt]

I = \int \frac{\sqrt{a^{2} (\sec^{2} t - 1)}}{a^{4} \sec^{4} t} . a \sec t \tan t dt

I = \frac{1}{a^{2}} \int \frac{\tan^{2} t}{\sec^{3} t} dt = \frac{1}{a^{2}} \int \sin^{2} t \cos t dt

I = \frac{1}{a^{2}} \int \sin^{2} t d (\sin t) = \frac{1}{{3 a}^{2}} \sin^{3} t + c

I = \frac{1}{{3 a}^{2}} {(\frac{\sqrt{x^{2} - a^{2}}}{x})}^{3} + c = \frac{1}{{3 a}^{2} x^{3}} \sqrt{{(x^{2} - a^{2})}^{3}} + c

Answered by bobhans last updated on 03/Sep/20

(★) L = \lim_{n \to \infty} [\frac{1}{n^{6}} \underset{k = 1}{\sum^{n}} {(n + k)}^{5}]

L = \lim_{n \to \infty} [\underset{k = 1}{\sum^{n}} {(1 + \frac{k}{n})}^{5} . \frac{1}{n}]

recall \underset{a}{\int^{b}} f (x) dx = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} f (x + k . △ x) . △ x

where △ x = \frac{b - a}{n}, we have △ x = \frac{1}{n}

by letting \to {\begin{cases} a = 1 \\ b = 2 \end{cases}

therefore L = \underset{1}{\int^{2}} x^{5} dx = {[\frac{1}{6} x^{6}]}_{1}^{2} = \frac{63}{6} = \frac{21}{2}

Commented by bemath last updated on 03/Sep/20

yes \dots thank you

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

\lim_{n \to \infty} \frac{1}{n} ({(1 + \frac{1}{n})}^{5} + {(1 + \frac{2}{n})}^{5} + \dots . n)

\lim_{n \to \infty} \frac{1}{n} \underset{r = 1}{\sum^{n}} {(1 + \frac{r}{n})}^{5}

= \int_{0}^{1} {(1 + x)}^{5} d x = \frac{1}{6} (64 - 1) = \frac{21}{2}