Convert $\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{4}{n}\sqrt{2+\frac{4k}{n}}$ into a definite integral.

Question

$$\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{4}{n}\sqrt{2+\frac{4k}{n}}$$ So going through this, I found out $a=2$ and $\Delta x = 4$, which got me $ \int_2^6 \sqrt{2+\frac {4k} n} dx $. If this is correct, how do I then evaluate this?

Answer 1

In this case $\Delta x = \frac{b-a}{n}$ and $x_k = a+k \Delta x$ with $a=2$, so $b=6$ and $x_k = 2 + \frac{4k}{n}$.

Answer 2

If you do some rearrangements, you get $$\lim_{n \to \infty} \frac{4}{n} \sum_{k=0}^n \sqrt{2 + 4\frac kn}$$ This is a riemann sum of the form $\lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n} f(\frac kn)$. The correct definite integral would be $\int_0^1 f(x) dx$, which in this case is $$4\int_0^1 \sqrt{2 + 4x} dx$$ Evaluating this integral is simple. Can you take it from here?