If $\Lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n-1} k \left[ \int_0^{k/n} \sqrt{(x-k)(k+1-x)} dx \right] = \frac{\pi}{m^n}$, then:
Step-by-Step Solution
Key Concept: The integral $\int_0^{k/n} \sqrt{(x-k)(k+1-x)} dx$ evaluates to $\frac{\pi}{8}$ using trigonometric substitution $x = k + \sin^2\theta$, transforming it to a Riemann sum that converges to $\int_0^1 x \cdot \frac{\pi}{8} dx = \frac{\pi}{16}$, making the limit equal $\frac{\pi}{16}$.
Given $x = k\cos^2\theta + (k+1)\sin^2\theta$, we have $x = k + \sin^2\theta$ and $dx = \sin 2\theta d\theta$. The integral $I = \int_0^{\pi/2} \sin(6\theta)2\sin\theta\cos\theta d\theta = \frac{1}{2}\int_0^{\pi/2}\sin^2 2\theta d\theta = \frac{\pi}{8}$ using substitution $2\theta = t$. Using Riemann sum: $L = \frac{1}{n^2}\sum_{k=0}^{n-1}k\left(\frac{\pi}{8}\right) = \frac{\pi(n-1)n}{16n^2} \to \frac{\pi}{16}$ as $n \to \infty$.
Correct Answer: 1,2