let-U-n-1-n-1-x-2-3-n-dx-calculate-lim-n-U-n-

Question Number 55214 by maxmathsup by imad last updated on 19/Feb/19

l e t U_{n} = \int_{\frac{1}{n}}^{1} \sqrt{x^{2} + \frac{3}{n}} d x . c a l c u l a t e {l i m}_{n \to + \infty} U_{n}

Commented by maxmathsup by imad last updated on 19/Feb/19

w e h a v e U_{n} = \int_{R} \sqrt{x^{2} + \frac{3}{n}} χ_{] \frac{1}{n}, 1]} (x) d x = \int_{R} f (n) d x w i t h

f_{n} (x) = \sqrt{x^{2} + \frac{3}{n}} χ_{] \frac{1}{n}, 1]} (x) d x t h e s e q u e n c e o f f u n c t i o n s f_{n} (x) v e r i f y

f_{n} (x) \to^{c s} x o n] 0, 1] a n d ∣ f_{n} (x) ∣= f_{n} (x) ⩽ \sqrt{x^{2} + 1} \forall x \in] 0, 1] t h e o r e m o f

c o n v e r g e n c e d o m i n e e g i v e {l i m}_{n \to + \infty} \int_{R} f_{n} (x) d x

= \int_{R} {l i m}_{n \to + \infty} f_{n} (x) d x = \int_{0}^{1} x d x = {[\frac{x^{2}}{2}]}_{0}^{1} = \frac{1}{2} \Rightarrow {l i m}_{n \to + \infty} U_{n} = \frac{1}{2} .

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19

\int_{\frac{1}{n}}^{1} \sqrt{x^{2} + {(\sqrt{\frac{3}{n}})}^{2}} d x

f o r m u l a

\int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} l n (x + \sqrt{x^{2} + a^{2}})

=∣ \frac{x}{2} \sqrt{x^{2} + \frac{3}{n}} + \frac{3}{n \times 2} l n (x + \sqrt{x^{2} + \frac{3}{n}}) ∣_{\frac{1}{n}}^{1}

= [{\frac{1}{2} \sqrt{1 + \frac{3}{n}} + \frac{3}{2 n} l n (1 + \sqrt{1 + \frac{3}{n}}} - {\frac{1}{2 n} \sqrt{\frac{1}{n^{2}} + \frac{3}{n}} + \frac{3}{2 n} l n (\frac{1}{n} + \sqrt{\frac{1}{n^{2}} + \frac{3}{n}})]

w h e n n \to \infty

\frac{1}{2} \times 1 = \frac{1}{2}