Medium MCQ +4 / -1 PYQ · JEE Mains 2021

The value of

$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}} $$ is :

  1. A ${1 \over 2}{\tan ^{ - 1}}(2)$
  2. B ${1 \over 2}{\tan ^{ - 1}}(4)$ Correct answer
  3. C ${\tan ^{ - 1}}(4)$
  4. D ${1 \over 4}{\tan ^{ - 1}}(4)$

Solution

$$L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}} $$<br><br>$\Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx}$<br><br>$$ \Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Rightarrow L = {1 \over 2}{\tan ^{ - 1}}4$$

About this question

Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals

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