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

Question

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

My try: $$\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}=\lim _{n\rightarrow +\infty} \frac{1}{n} \sum _{k=0}^{n} \sqrt{2+\frac{k}{n}-(\frac{k}{n})^2}=\lim _{n\rightarrow +\infty} \frac{1}{n} \sum _{k=0}^{n} f(\frac{k}{n})=\int^{1}_{0} \sqrt{2+x-x^2} dx=\int^{1}_{0} \sqrt{-(x-\frac{1}{2})^2+\frac{9}{4}} dx=\int^{\frac{1}{2}}_{-\frac{1}{2}} \sqrt{\frac{9}{4}-u^2} du$$In this sollution: $$f:[0,1]\rightarrow \mathbb R, f(x)=\sqrt{2+x-x^2}$$ $$u=x-\frac{1}{2}, du=dx$$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 $ which is problematic.

Can you help me how to bypass this problem?

Answer 1

Hint:

Use the substitution:

$$u = \frac{3}{2} \sin(t)$$

Answer 2

Hint. Just make the change of variable $$ u=\frac 32 \sin x,\qquad du = \frac 32 \cos x \,dx, $$ giving $$ \int^{\frac{1}{2}}_{-\frac{1}{2}} \sqrt{\frac{9}{4}-u^2} _,du=\frac 94\int^{\arcsin\frac{1}{3}}_{-\arcsin\frac{1}{3}} \cos^2 x \,dx=\frac 98\int^{\arcsin\frac{1}{3}}_{-\arcsin\frac{1}{3}}\left(1+ \cos 2 x\right) dx. $$ The conclusion is then standard.