Based on what you've said, I think a good place to start is at the very beginning, by reviewing the definition of a Riemann Sum. In essence, you're being asked to find the area between the given curve and the x-axis. The easiest way is, as you noted, to take the integral. However, we can get an approximation without integrating. The roughest possible approximation might be to take the rectangle created by the lines y = 0, y = 36, x = 0, and x =6. Such a rectangle has height 36 and width 6, for an area of 216. But clearly that's too much because a graphical inspection shows large swatches of area covered by this rectangle that aren't underneath the curve. Hm... well, what if we split the curve into two rectangles by drawing a vertical line at x = 3? Those two rectangles would more closely approximate the area underneath the curve a little better than did one rectangle. Adding a third rectangle would make the approximation even tighter and so on. I inadvertently graphed the function from x = -6 to x = 6, but this diagram may give some insight into what's going on. Here, the area is subdivided into 12 rectangles.
The Riemann Sum, then, gives a mathematical way of describing the total area of these rectangles. We've noticed that as the number of rectangles increases (and each rectangle therefore has smaller and smaller width), the approximation becomes closer to the actual area, as given by the integral. Indeed, the limit of this Riemann Sum, as we approach infinitely many rectangles with infinitely small width, does give the exact value. The formula for a Riemann Sum (hopefully you've seen this in your class notes and/or textbook) of some function f(x) over some interval [a, b] is given by:
\(\displaystyle \displaystyle \lim_{n \to \infty} \left( \sum_{k=1}^{n} f(x_k) \cdot \Delta x_k \right)\)
There are many different types of Riemann Sums, each with their own unique rules for what \(\displaystyle x_k\) is, but one of the easiest ones to use is what's called a "right sum," in which \(\displaystyle x_k\) is defined as the right-most point in each rectangle. But no matter which Riemann Sum you use, \(\displaystyle \Delta x_k\) always represents the width of the rectangles and is given by:
\(\displaystyle \Delta x_k = \dfrac{b-a}{n}\)
Using the right sum definition, each x_k would simply be one further increment of the width of a rectangle along the curve. For the case with 24 rectangles (i.e. n = 24), the values of x_k would therefore be: 0, 1/4, 2/4, 3/4, 1, ..., 6. In other words, the general \(\displaystyle x_k = k \cdot \Delta x_k\), making the full Riemann Sum:
\(\displaystyle \displaystyle \lim_{n \to \infty} \left( \sum_{k=1}^{n} f \left( \dfrac{k(b-a)}{n} \right) \cdot \dfrac{b-a}{n} \right)\)
What happens when you evaluate this Riemann Sum for your given f(x)? Where does that lead you?
