$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2}$ the problem of $k = 0$. Am I doing it right? I would like to calculate:
$$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2}$$
We have that:
$$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^n \sqrt{2+\frac{k}{n}-(\frac{k}{n})^2} $$
Now, since we have: $\lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^n f(\frac{k}{n})$ I would like to use the fact that:
$$
\lim_{n \to \infty} \frac{b-a}{n} \sum^n_{k=1}f(a + k \frac{b-a}{n})= \int^b_a f(x) \ dx,
$$
But here I have $k = 0$ and not $k = 1$. So do I have to put $l = k+1 \to k = l-1$?
In that case we have:
$$\lim_{n \to \infty} \frac{1}{n}\sum_{l=1}^n \sqrt{2+\frac{l+1}{n}-(\frac{l+1}{n})^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{l=1}^n \sqrt{2 - \frac{l}{n}-\frac{l^2}{n}} $$
Now I can use the equation:
$$\lim_{n \to \infty} \frac{1}{n}\sum_{l=1}^n \sqrt{2 - \frac{l}{n}-\frac{l^2}{n}}  = \int^1_0 \sqrt{2 - \frac{l}{n}-\frac{l^2}{n}} \ dx$$
Is that correct up to this point? Am I allowed to do change $k$ for $l$ like that?

No.
$$\sum_{k=0}^n\frac{\sqrt{2n^2+kn-k^2}}{n^2}=$$
$$\frac{\sqrt{2}}{n}+\sum_{k=1}^n\frac{\sqrt{2n^2+kn-k^2}}{n^2}$$
the first term goes to zero and the second to
$$\int_0^1\sqrt{2+x-x^2}dx=$$
$$\int_0^1\sqrt{\frac 94-(x-\frac 12)^2}dx$$
now put
$$x-\frac 12=\frac 32\sin(t)$$
