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Calculus  II                                                          Quiz #4                                                March 11, 2003

Name____________________                      R.  Hammack                                              Score ______
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(1) Find the area contained between the graphs of f(x) = 6 - x  and   g(x) = x^2 - 2x.

First, let's graph these two functions to see what we are dealing with. The graph of f is a straight line with y-intercept 6 and x-intercept 6. Factoring g, we get  g(x) = x^2 - 2x = x(x - 2), so you can see that the graph of g is a parabola that opens "up" and has x-intercepts 0 and 2, and y-intercept 2. Here is a picture.

[Graphics:HTMLFiles/quiz4_4.gif]
So the region we are interested in is the crescent-moon shape contained between the graph of f (on top) and g (on the bottom). To find the x values that bound the region on the left and right, we need to solve to find the intersection points.

f(x) = g(x)
6 - x = x^2 - 2x
0 = x^2 - x - 6
0 = (x - 3) (x + 2)

This means the two graphs intersect at the x values of -2 and 3.
Thus the area of the region is

∫_ (-2)^3 (f(x) - g(x)) dx =∫_ (-2)^3 ((6 - x) - (x^2 - 2x)) dx =∫_ (-2)^3 (-x^2 + x + 6) dx =[-x^3/3 + x^2/2 + 6x] _ (-2)^3

= (-3^3/3 + 3^2/2 + 6 (3)) - (-(-2)^3/3 + (-2)^2/2 + 6 (-2))

= (-9 + 9/2 + 18) - (8/3 + 2 - 12)

= 9 + 9/2 - 8/3 + 10 =

= 54/6 + 27/6 - 16/6 + 60/6 = 125/6 square units.