Calculus, integration, Riemann sum help? Express as a deﬁnite integral and then evaluate the limit of the Riemann sum lim
$$
\lim_{n\to \infty}\sum_{i=0}^{n-1} (3x_i^2 + 1)\Delta x,
$$
where $P$ is the partition with
$$
x_i = -1 + \frac{3i}{n}
$$
for $i = 0, 1, \dots, n$ and $\Delta x \equiv x_i - x_{i-1}$.
I am completely and utterly confused as to how to even start this question. Any help/good links hugely appriciated!
 A: When $i$ goes from $0$ to $n$, then $-1+\dfrac{3i}{n}$ goes from $-1$ to $2$.  There you have the bounds of integration.  What you're integrating is $3x^2+1$.  So evaluate that integral between those bounds.
A: Let $f$ be a function, and let $[a,b]$ be an interval. Let $n$ be a positive integer, and let $\Delta x=\frac{b-a}{n}$. Let $x_0=a$, $x_1=a+\Delta x$, $x_2=a+2\Delta x$, and so on up to $x_n=a+n\Delta x$.  So $x_i=a+i\Delta x$.
So far, a jumble of symbols. You are likely not to ever understand what's going on unless you associate a picture with these symbols. 
So draw some nice function $f(x)$, say always positive, and take some interval $[a,b]$.  For concreteness, let $a=1$ and $b=4$. Pick a specific $n$, like $n=6$. Then $\Delta x=\frac{3}{6}=\frac{1}{2}$.
So $x_0=1$, $x_1=1.5$, $x_2=2$, $x_3=2.5$, $x_4=3$, $x_5=2.5$, and $x_6=3$. Note that the points divide the interval from $a$ to $b$ into $n$ subintervals. These intervals all have width $\Delta x$.  
Now calculate $f(x_0)\Delta x$. This is the area of a certain rectangle. Draw it. Similarly, $f(x_1)\Delta x$ is the area of a certain rectangle. Draw it. Continue up to $f(x_5)\Delta x$.  Add up. The sum is called the left Riemann sum associated with the function $f$ and the division of $[1,4]$ into $6$ equal-sized parts.
The left Riemann sum is an approximation to the area under the curve $y=f(x)$, from $x=a$ to $x=b$. Intuitively, if we take $n$ very large, the sum will be a very good approximation to the area, and the limit as $n\to\infty$ of the Riemann sums is the integral $\displaystyle\int_a^b f(x)\,dx$.
Let us apply these ideas to your concrete example. It is basically a matter of pattern recognition. We have $x_0=-1$, $x_1=-1+\frac{3}{n}$, $x_2=-1+\frac{6}{n}$, and so on. These increase by $\frac{3}{n}$, so $\Delta x=\frac{3}{n}$. 
We have $x_0=-1$, and $x_n=-1+\frac{3n}{n}=2$. So $a=-1$ and $b=2$.
Our sum is a sum of terms of the shape $(3x_i^2+1)\Delta x$. Comparing with the general pattern $f(x_i)\Delta x$, we see that $f(x)=3x^2+1$.
So for large $n$, the Riemann sum of your problem should be a good approximation to $\displaystyle\int_{-1}^2 (3x^2+1)\,dx$. 
