Section 6.1 Worksheet:

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

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.