Calcular

Question

Calcular $$\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}$ $

Mi intento: $$\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}=\lim _{n\rightarrow +\infty} \frac{1}{n} \sum _{k=0}^{n} \sqrt{2+\frac{k}{n}-(\frac{k}{n})^2}=\lim _{n\rightarrow +\infty} \frac{1}{n} \sum _{k=0}^{n} f(\frac{k}{n})=\int^{1}_{0} \sqrt{2+x-x^2} dx=\int^{1}_{0} \sqrt{-(x-\frac{1}{2})^2+\frac{9}{4}} dx=\int^{\frac{1}{2}}_{-\frac{1}{2}} \sqrt{\frac{9}{4}-u^2} du$ $ En esta solución: $$f:[0,1]\rightarrow \mathbb R, f(x)=\sqrt{2+x-x^2}$ $ $$u=x-\frac{1}{2}, du=dx$$Unfortunatelly I don't know what I can do with $ \ int ^ {\ frac {1} {2}} _ {- \ frac {1} {2}} \ sqrt {\ frac {9} {4} -u ^ 2} du$ because my only idea is integration by substitution. But if I use for example $ s = u ^ 2$ then I have $ ds = 2udu$ so $ du = \ frac {ds } {2u}$ so I did not get rid of $ u $ lo cual es problemático.

¿Puedes ayudarme a evitar este problema?

Answer 1

Insinuación:

Utilice la sustitución:

PS

Answer 2

Sugerencia Simplemente haga el cambio de la variable $$ u = \ frac 32 \ sin x, \ qquad du = \ frac 32 \ cos x \, dx, $$ dando $$ \ int ^ {\ frac {1} {2}} _ {- \ frac {1} {2}} \ sqrt {\ frac {9} {4} -u ^ 2} _, du = \ frac 94 \ int ^ {\ arcsin \ frac {1} {3}} _ {- \ arcsin \ frac {1} {3}} \ cos ^ 2 x \, dx = \ frac 98 \ int ^ {\ arcsin \ frac {1} {3}} _ {- \ arcsin \ frac {1} {3}} \ left (1+ \ cos 2 x \ right) dx. $$ La conclusión es entonces estándar.