<s>[INST] Calculate $\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}$ 
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?
 [/INST] 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.
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