Evaluate $\lim_{n \rightarrow \infty} \frac{1}{n} [(1/n)^2+({2}/{n})^2+ \cdots +({3n}/{n})^2]$ $$\lim_{n \rightarrow \infty} \frac{1}{n} \Bigg[\Bigg(\frac{1}{n}\Bigg)^2+\Bigg(\frac{2}{n}\Bigg)^2+ \cdots +\Bigg(\frac{3n}{n}\Bigg)^2\Bigg]$$

Express this sum as an integral. I know $\Delta x = \frac{1}{n}$ and $f(x) = x^2$.

$$\lim_{n \rightarrow \infty} \sum_{k=1}^{3n}\frac{1}{n} \Big(\frac{k}{n}\Big)^2$$ The $3n$ is tripping me up. I know $$\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{1}{n} \Big(\frac{k}{n}\Big)^2 = \int_{0}^{1}x^2 dx$$ 
Any help is appreciated.
Hint: Break the sum into three parts. 

Here is the second part:

$$\frac1{n} \sum_{k= n+1}^{2n}\left(\frac{k}{n}\right)^2= \frac1{n} \sum_{k= 1}^{n}\left(\frac{n+k}{n}\right)^2=\frac1{n} \sum_{k= 1}^{n}\left(1 + \frac{k}{n}\right)^2\\ \to\int_0^1(1+x)^2\,dx = \int_1^2x^2 \, dx$$
Hint: What is $\frac{3n}{n}$?

Everything else looks good.