Definite Integrals

Question

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

RankGuru · Accepted Answer

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: $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(x/a)$.
$I = [\sin^{-1}(x/2)]_0^1 = \sin^{-1}(1/2) - \sin^{-1}(0)$.
$I = \pi/6 - 0 = \pi/6$.

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