Determine which of the following is equal to $$ \lim_{n\to \infty} \frac{34}{n} \sum_{i=1}^n \left(\frac{34^2i^2}{n^2} + 1 \right) $$ (Answer: the area of the region above the $x$-axis below $f(x)=(x-1)^2+1$, from $x=1$ to $x=35$.)
I got a correct answer only because I knew the final $x$ and the initial $x$ have to have a difference of $34$. I have no idea how this function came about though and I would like some help
The form of the sum suggests you are adding $n$ rectangles of width $34/n$ and different heights given by the sum. The common width, as you mention, suggests that you have $n$ rectangles over an interval of width $34$, which we can represent as $(a,a+34)$.
As for the heights, the height of rectangle $i$, over the interval $\left( a+\frac{34(i-1)}{n}, a+\frac{34i}{n}\right)$ is given by the $i$-th term of the sum, namely $\frac{34^2i^2}{n^2}+1$, which sounds a lot like one of those endpoints plugged into the function $f(x)=x^2+1$.
Can you now finish?
UPDATE
Imagine writing down a Riemann sum of $n$ rectangles for $\int_a^{a+L} \left(x^2+1\right) dx$. The width of one rectangle is $L/n$, and the $i$-th interval is $(a + (i-1)L/n, a+iL/n)$, so if you are using the RHS endpoint the height will be $(a+iL/n)^2+1$ and if you want the LHS endpoint, you get $(a + (i-1)L/n)+1$.
So the total sum for your integral becomes $$ \int_a^{a+L} \left(x^2+1\right)dx = \lim_{n \to \infty} \frac{L}{n} \sum_{i=1}^n \left[\left(a+\frac{iL}{n}\right)^2+1\right] $$