TITLE: Convert Riemann sum to a definite integral: $\lim_{n\to \infty} \sum_{k=1}^n\sqrt{{1\over n^2}\left(1+{2k\over n}\right)} $
QUESTION [5 upvotes]: I'm between 2 answers for this question, but I am not sure if either of them are right.
$$\lim_{n\to \infty} \sum_{k=1}^n\sqrt{{1\over n^2}\left(1+{2k\over n}\right)} $$
It has to be rewritten as a definite integral. I am between $$\int_0 ^1\sqrt{1+2x}dx$$ and $$\int_1^2\sqrt{x}dx$$ Not sure if either of them or right though, help please.

REPLY [2 votes]: One may recall that for any Riemann integrable function over $[a,b]$, as $n \to \infty$, one has

$$
\sum_{k=0}^n\frac{(b-a)}nf\left(a+\frac{k(b-a)}n \right) \to \int_a^bf(x)dx
$$

Then you may apply it to $\displaystyle f(x)=\sqrt{1+2x}$, $a=0$, $b=1$ giving the limit
$$
\int_0 ^1\sqrt{1+2x}\:dx.
$$ By the change of variable $1+2x \to u$ you also get
$$
\frac12\int_1^3\sqrt{u}\:du
$$