First slide

Introduction to definite integration

Question

The $lim_{n \to \infty} S_{n}$ if $S_{n} = \frac{1}{2 n} + \frac{1}{\sqrt{4 n^{2} - 1}} + \frac{1}{\sqrt{4 n^{2} - 4}} + \dots + \frac{1}{\sqrt{3 n^{2} + 2 n - 1}}$ is

Easy

A

$π / 2$
B

2
C

1
D

$π / 6$

Solution

$lim_{n \to \infty} S_{n}$
$\begin{array}{l} = lim_{n \to \infty} \frac{1}{n} [\frac{1}{\sqrt{4 - 0}} + \frac{1}{\sqrt{4 - 1 / n^{2}}} + \frac{1}{\sqrt{4 - 4 / n^{2}}} + \dots + \sqrt{4 - {(\frac{n - 1}{n})}^{2}}] \\ = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} \frac{1}{\sqrt{4 - (r / n)^{2}}} = \int_{0}^{1} \frac{d x}{\sqrt{4 - x^{2}}} = {\sin^{- 1} \frac{x}{2}|}_{0}^{1} = \frac{π}{6} . \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App