Section 6.1 Worksheet:

Assigned problems: Exercises pp. 376-378, #2, 3, 9, 20, 35, 40, 41, 44 (due Friday, 4/4).

1. An integral doesn't really represent an area, as the Greeks understood area, because an integral can be negative (and area can't). So we consider integrals as sums of positive and ``negative'' areas. If you actually want the physical, Greek-style area between two curves, what formula must you use?

2. Illustrate the situation described above with a sketch.

3. One twist in this section is that we sometimes have functions of y, rather than x. Draw an example, and explain how you must proceed differently (if at all!).

Notes:

1. This section is a simple generalization of the definition of a definite integral as a sum of positive and negative areas bounded by

   • the graph of a function,
   • the x-axis,
   • and two vertical lines at x=a and x=b.
   Now we simply consider areas bounded by
   • the graph of a function,
   • the graph of a second function,
   • and two vertical lines at x=a and x=b.
   The consequence is simply that we need to do the difference of two integrals, rather than a single integral. Twice the work, with scarcely any additional complexity.