The value of $\lim_{n \to \infty} \sum_{r=1}^n \frac{1}{\sqrt{4n^2 - r^2}}$ is:

1. Express the limit as a Riemann Sum: Term $= \frac{1}{\sqrt{4n^2 - r^2}} = \frac{1}{n \sqrt{4 - (r/n)^2}} = \frac{1}{n} \frac{1}{\sqrt{4 - (r/n)^2}}$. 2. Convert to integral: Let $x = r/n, dx = 1/n$. Limits from $0$ to $1$. Integral $= \int_0^1 \frac{1}{\sqrt{2^2 - x^2}} dx$. 3. Solve the integral: Standard formula: $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(x/a)$. $I = [\sin^{-1}(x/2)]_0^1 = \sin^{-1}(1/2) - \sin^{-1}(0)$. $I = \pi/6 - 0 = \pi/6$.

Definite Integrals

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The value of $lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{4 n ^{2} - r ^{2}}$ is:

Select the correct option:

$π /6$

$π /2$

$π /3$

$π /4$

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About This Question

Subject: mathematics
Chapter: integral calculus
Topic: definite integrals
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

$π /6$

Express the limit as a Riemann Sum: Term $= \frac{1}{4 n ^{2} - r ^{2}} = \frac{1}{n 4 - ( r / n ) ^{2}} = \frac{1}{n} \frac{1}{4 - ( r / n ) ^{2}}$ .
Convert to integral: Let $x = r / n, d x = 1/ n$ . Limits from $0$ to $1$ . Integral $= \int_{0}^{1} \frac{1}{2 ^{2} - x ^{2}} d x$ .
Solve the integral: Standard formula: $\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} (x / a)$ . $I = [sin^{- 1} (x /2)]_{0}^{1} = sin^{- 1} (1/2) - sin^{- 1} (0)$ . $I = π /6 - 0 = π /6$ .

This medium difficulty mathematics question is from the chapter integral calculus, covering the topic of definite integrals. It appeared in the 2025 exam.

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