66We have
(x-k)(k+1-x)=\left(\dfrac{1}{2}\right)^2-\left(x-\dfrac{2k+1}{2}\right)^2
Hence,
\int_k^{k+1}\sqrt{(x-k)(k+1-x)}\ \mathrm dx=\int_k^{k+1}\sqrt{\left(\dfrac{1}{2}\right)^2-\left(x-\dfrac{2k+1}{2}\right)^2}\ \mathrm dx
Apply the substitution
x-\dfrac{2k+1}{2}=\dfrac{1}{2}\sin\theta\quad\Rightarrow\ \mathrm dx=\dfrac{1}{2}\cos\theta \ \mathrm d\theta
Also when x=k, sinθ = -1, i.e. θ=-π/2 and when x=k+1, sinθ=1, i.e. θ=π/2. Hence the given integral is \int_{-\pi/2}^{\pi/2}\dfrac{1}{4}\cos^2\theta \ \mathrm d\theta =\dfrac{\pi}{8}
Hence, the required limit becomes
\lim_{n\to\infty}\dfrac{1}{n^2}\sum_{k=0}^{n-1}\dfrac{k\pi}{8}=\lim_{n\to\infty}\dfrac{1}{n^2} \dfrac{\pi}{8}\dfrac{n(n-1)}{2}=\dfrac{\pi}{16}
