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                                   EXAMPLE 5.20         CHAPTER 5         Indeterminate Forms
                                   Calculate the area above the x-axisand under the curve
                                                            1
                                                    y =     4/3  ,   0 <x ≤ 1/2.
                                                        x · ln  x
                                   SOLUTION
                                     According to the preceding discussion, this area is equal to the value of the
                                   improper integral

                                                 1/2   1                 1/2   1
                                                            dx = lim                 dx.
                                               0   x · ln 4/3  x  (→0 +  (  x · ln 4/3  x

                                   For clarity we let ϕ(x) = ln x, ϕ (x) = 1/x. Then the (indefinite) integral
                                   becomes

                                                        ϕ (x)           3
                                                              dx =−         .
                                                       ϕ 4/3 (x)      ϕ 1/3 (x)
                                   Thus

                                                                                1/2
                                               1/2    1                    3
                                         lim                dx = lim −
                                        (→0 +  (  x · ln 4/3  x  (→0 +  ln  1/3
                                                                             x (

                                                                          −3          −3
                                                               = lim             −          .
                                                                 (→0 +  [− ln 2] 1/3  [ln (] 1/3
                                                                            1/3
                                   Now as ( → 0 then ln ( →−∞ hence 1/[ln (]   → 0. We conclude that our
                                   improper integral converges and the area under the curve and above the x-axis
                                                1/3
                                   equals 3/[ln 2]  .




                   5.4 More on Improper Integrals



                               5.4.1      INTRODUCTION
                               Suppose that we want to calculate the integral of a continuous function f(x) over
                               an unbounded interval of the form [A, +∞) or (−∞,B]. The theory of the integral
                               that we learned earlier does not cover this situation, and some new concepts are
                               needed. We treat improper integrals on infinite intervals in this section, and give
                               some applications at the end.
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