Page 155 - Calculus Demystified
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                                   SOLUTION             CHAPTER 5         Indeterminate Forms
                                     We do this problem by evaluating the limit
                                                   −32                   5
                                                                                −32
                                            lim        x −1/5  dx =  lim  x 4/5
                                           M→−∞   M               M→−∞ 4        M
                                                                        5 
     4/5    4/5
                                                               =   lim     (−32)   − M
                                                                  M→−∞ 4
                                                                        5 
       4/5
                                                               =   lim     16 − M    .
                                                                  M→−∞ 4
                                   This limit equals −∞. Therefore the integral diverges.

                                                      ∞
                               You Try It: Evaluate   (1 + x) −3  dx.
                                                    1
                                  Sometimes we have occasion to evaluate a doubly infinite integral. We do so by
                               breaking the integral up into two separate improper integrals, each of which can be
                               evaluated with just one limit.
                                   EXAMPLE 5.23
                                   Evaluate the improper integral

                                                              ∞
                                                                   1
                                                                       dx.
                                                                 1 + x 2
                                                             −∞
                                   SOLUTION
                                     The interval of integration is (−∞, +∞). To evaluate this integral, we break
                                   the interval up into two pieces:

                                                   (−∞, +∞) = (−∞, 0]∪[0, +∞).
                                   (The choice of zero as a place to break the interval is not important; any
                                   other point would do in this example.) Thus we will evaluate separately the
                                   integrals
                                                                         0
                                                   +∞     1                  1
                                                              dx and             dx.
                                                  0    1 + x 2         −∞ 1 + x 2
                                   For the first one we consider the limit
                                                    N   1                        N

                                             lim            dx =   lim  Tan −1
                                                                              x
                                           N→+∞    0  1 + x 2    N→+∞           0

                                                               =   lim   Tan −1 N − Tan −1 0
                                                                 N→+∞
                                                                  π
                                                               =   .
                                                                  2
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