Hi All,
Here are the solutions to these problems.
1. Please complete each of these geometrically - that is, sketch the graph and find the area using shapes that you know.
a. 
Here is a graph of the equation on the interval in question (marked by pink dashed lines). If we are looking for the area, we can divide this into a rectangle (width 3, height 1), and a triangle (width 3, height 9). Combining these two areas, I get 3+27/2=33/2 or 16.5
b. 
Here is a graph of the equation on the interval in question (marked by pink dashed lines). Here we have a positive area (above the x-axis, carefully identified in green) and a negative area (below the x-axis, carefully identified in blue). We need to find the area of each triangle, and add them together. The line crosses the x-axis at (1.5,0), so that is where we will stop. The positive triangle has a height of 5 (plug -1 into the equation), and width of 2.5. Its area will be 6.25. The negative triangle has a height of 7 and a width of 3.5. It’s area will be -12.25. Combining the two areas: 6.25 + -12.25 = -6.
3. Explain how to find 
using antiderivatives.
First take the antiderivative of the function: -x^2+3x+C.
Plug in 5 (-25+15+C), plug in -1 (-(1)+-3+C) to get (-10+C) and (-4+C) respectively, and subtract them.
-10+C - -4 -C = -6
You got the same answer using area as you did by doing the antiderivative.