When you sketch the graph, you will see the region in question is a right triangle in QI with vertices at the origin, at (0 , 12) (the y-int of the line), and at (4 , 0) (the x-int of the line). The base, 0 ≤ x ≤ 4, is the x-interval over which we are doing the Riemann sum. We will create 4 rectangles of equal width (so each is 1 unit wide), and whose heights will be determined by y-coordinates on the line y = 12 - 3x. A "lower" sum means in this case using the right-hand endpoints of the subintervals, which are x = [0 , 1] , [1 , 2] , [2 , 3] , and [3 , 4]. So the four x-values where we will find rectangle heights are x = 1, 2, 3, and 4.
All we need to do now is draw those 4 rectangles, get their heights, and sum their areas:
When x = 1, y = 9, when x = 2, y = 6, when x = 3, y = 3, and when x = 4, y = 0.
So the sum of the 4 rectangular areas is this: A = 9·1 + 6·1 + 3·1 + 0·1 = 18.
Of course, for a linear function such as this one, approximating the area with rectangles seems a pretty silly exercise: we can calculate the area under this "curve" exactly by using a geometry formula. However, if you instead apply the same process to a function with a curved graph, say y = 12 - .75x2, you could better see why using rectangles to approximate that area would be helpful. (That is, as long as you thought that calculating that area would have some practical application / interpretation we cared about, which it turns out it very much does.)
