7 In this exercise, we will compute the denite integral / | 33(9: 5)| dm. 0 Yes, that's an absolute value in the integral, and we haven't covered those in the lectures. But we'll convince you that you can absolutely do this. And when you come up with the main insight that solves this problem, you'll pretty much discovered on your own how to handle absolute values in integrals in general. 7 Here is an idea that can help: Think in terms of area! The integral [0 l 3m(m 5)| dm corresponds to the area underneath the curve of (the nonnegative function) | 322(2: 5)| over [0, 7], as shown in the graph below. Now, stare at the graph, think of how you can possibly break down the problem, and give it a try! L7I3m(m5)|dm=