[tex]\displaystyle l=\lim\limits_{n\to \infty}\dfrac{1}{n^3}\sum\limits_{k=1}^{2n}\dfrac{k^2}{8}\arcsin\dfrac{k}{2n}\\ \\2n = t\Rightarrow n = \dfrac{t}{2}\Rightarrow t\to \infty \\ \\ l = \lim\limits_{t\to \infty} \dfrac{1}{\Big(\dfrac{t}{2}\Big)^3}\sum\limits_{k=1}^{t}\dfrac{k^2}{8}\arcsin\dfrac{k}{t} \\ \\ l =\lim\limits_{t\to \infty}\dfrac{8}{t^3}\sum\limits_{k=1}^{t}\dfrac{k^2}{8}\arcsin\dfrac{k}{t}\\ \\ l = \lim\limits_{t\to \infty}\dfrac{1}{t^3}\sum\limits_{k=1}^{t}k^2}\arcsin\dfrac{k}{t}[/tex]
[tex]\displaystyle l = \lim\limits_{n\to \infty}\dfrac{1}{n^3}\sum\limits_{k=1}^{n}k^2}\arcsin\dfrac{k}{n}\\ \\ l =\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{k=1}^{n}\dfrac{k^2}{n^2}}\arcsin\dfrac{k}{n}\\ \\ l = \lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{k=1}^{n}\Big(\dfrac{k}{n}}\Big)^2\arcsin\Big(\dfrac{k}{n}\Big)[/tex]
Se observă că este o sumă Riemann.
Transformând-o în integrală Riemann obținem:
[tex]\displaystyle l = \int_{0}^1(x^2\arcsin x)\, dx = \dfrac{x^3\arcsin x}{3}\Big|_{0}^1-\dfrac{1}{3}\int_{0}^1\dfrac{x^3}{\sqrt{1-x^2}}\, dx \\ \\ \\\sqrt{1-x^2} = t \Rightarrow 1-x^2 = t^2 \Rightarrow x^2 = 1-t^2\Rightarrow 2x\,dx = -2t\,dt\Rightarrow\\ \Rightarrow x\, dx = -t\, dt \\ \\ x = 0 \Rightarrow t = 1,\quad x = 1 \Rightarrow t = 0[/tex]
[tex]\displaystyle I_1 = \int_{0}^1\dfrac{x^2}{\sqrt{1-x^2}}\cdot x\, dx = \int_{0}^1\dfrac{1-t^2}{t}\cdot t\, dt = \int_{0}^1(1-t^2)\, dt = \\ \\ = \Big(t-\dfrac{t^3}{3}\Big)\Big|_{0}^1 = 1-\dfrac{1}{3} = \dfrac{2}{3}\\ \\ \\ l =\dfrac{x^3\arcsin x}{3}\Big|_{0}^1-\dfrac{1}{3}\cdot \dfrac{2}{3} =\dfrac{1\cdot \dfrac{\pi}{2}}{3}-\dfrac{2}{9} = \boxed{\dfrac{\pi}{6}-\dfrac{2}{9}}[/tex]
