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Question Number 63507 by mathmax by abdo last updated on 05/Jul/19

l e t U_{n} = \int_{\frac{1}{n}}^{1} (\sqrt{x^{2} + x + 1} - \sqrt{x^{2} - x + 1}) d x (n > 0)

1) c a l c u l a t e {l i m}_{n \to + \infty} U_{n}

2) f i n d n a t u r e o f Σ U_{n}

Answered by MJS last updated on 05/Jul/19

\lim_{n \to + \infty} U_{n} = \underset{0}{\int^{1}} (\sqrt{x^{2} + x + 1} - \sqrt{x^{2} - x + 1}) d x

\int \sqrt{x^{2} \pm x + 1} d x = \frac{1}{2} \int \sqrt{{(2 x \pm 1)}^{2} + 3} d x =

[t = \sinh^{- 1} \frac{2 x \pm 1}{\sqrt{3}} \to d x = \sqrt{x^{2} \pm x + 1} d t]

= \frac{3}{4} \int (\cosh^{2} t) d t = \frac{3}{8} \int (1 + \cosh 2 t) d t =

= \frac{3}{8} t + \frac{3}{16} \sinh 2 t

\int (\sqrt{x^{2} + x + 1} - \sqrt{x^{2} - x + 1}) d x =

= \frac{3}{8} (\sinh^{- 1} \frac{2 x + 1}{\sqrt{3}} - \sinh^{- 1} \frac{2 x - 1}{\sqrt{3}}) + \frac{1}{4} ((2 x + 1) \sqrt{x^{2} + x + 1} - (2 x - 1) \sqrt{x^{2} - x + 1}) + C

\lim_{n \to + \infty} U_{n} = \underset{0}{\int^{1}} (\sqrt{x^{2} + x + 1} - \sqrt{x^{2} - x + 1}) d x =

= - \frac{3 - 3 \sqrt{3}}{4} + \frac{3}{8} \sinh^{- 1} \sqrt{3} - \frac{9}{8} \sinh^{- 1} \frac{\sqrt{3}}{3} \approx . 424928

[= - \frac{3 - 3 \sqrt{3}}{4} + \frac{3}{8} \ln (3 + 2 \sqrt{3}) - \frac{3}{4} \ln 3]

Σ U_{n} = + \infty

Commented by mathmax by abdo last updated on 05/Jul/19

t h a n k y o u s i r m j s f o r t h o s e h a r d w o r k s .

Commented by MJS last updated on 05/Jul/19

{you}^{'} re welcome . you know I love integrals \dots

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