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                               If f (x) is negative over all or part of [a, b], the integral
                                 b

                                   f (x) dx will not give the desired area; the area must be de-
                                a
                                                          b

                               termined by computing       | f (x)| dx. Since the Fundamental
                                                         a
                               Theorem of Calculus cannot be conveniently applied to a
                               function involving an absolute value, one must first deter-
                               mine where the graph crosses the x axis, break up the interval
                               into subintervals determined by these points, integrate sepa-
                               rately in each subinterval, and add the absolute values of the
                               integrals. The next example illustrates this procedure.


                               EXAMPLE 3
                                                                              2
                               Find the area of the region bounded by y = x − 5x + 6, the
                               x axis, and the vertical lines x = 0 and x = 4.

                                   Solution
                                   First we draw a sketch of the area to be determined.
                                     6 6

                                     5 5

                                     4 4

                                     3 3

                                     2 2
                                     1 1


                                                  1 1         2 2        3 3         4 4

                               Next, we determine where the graph crosses the x axis.

                                                          2
                                                     y = x − 5x + 6
                                                          2
                                                     0 = x − 5x + 6
                                                     0 = (x − 2)(x − 3)

                                                     x = 2     x = 3

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