Page 109 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 109

100                                 INTEGRALS                              [CHAP. 5


                                                        n
                                                       X      2       2 1=2
                                                   lim   fð x k Þ þð y k Þ g
                                                       k¼1
                                                   n!1
                     or equivalently
                                                        (          ) 1=2
                                                      n            2
                                                     X         y k
                                                  lim    1 þ          ð x k Þ
                                                     k¼1
                                                  n!1          x k
                     where  x k ¼ x k   x k 1 and  y k ¼ y k   y k 1 .
                        Thus, the length of the arc of a curve in rectangular Cartesian coordinates is
                                                                   (            ) 1=2
                                            b                        dx     dy
                                           ð                      ð      2       2
                                                  2       2 1=2
                                                0      0                           dt
                                        L ¼  f½ f ðtÞ Šþ½g ðtފ g  dt ¼  þ
                                            a                        dt     dt
                        (This form may be generalized to any number of dimensions.)
                        Upon changing the variable of integration from t to x we obtain the planar form
                                                           (         ) 1=2
                                                        ð            2
                                                                 dy
                                                         f ðbÞ
                                                    L ¼      1 þ
                                                                 dx
                                                         f ðaÞ
                        (This form is only appropriate in the plane.)
                                                              2
                                                         2
                                                    2
                        The generic differential formula ds ¼ dx þ dy is useful, in that various representations algebrai-
                     cally arise from it.  For example,
                                                             ds
                                                             dt
                     expresses instantaneous speed.
                     AREA
                        Area was a motivating concept in introducing the integral. Since many applications of the integral
                     are geometrically interpretable in the context of area, an extended formula is listed and illustrated below.
                        Let f and g be continuous functions whose graphs intersect at the graphical points corresponding to
                     x ¼ a and x ¼ b, a < b.If gðxÞ A f ðxÞ on ½a; bŠ, then the area bounded by f ðxÞ and gðxÞ is

                                                         ð b
                                                           fgðxÞ  f ðxÞg dx
                                                     A ¼
                                                          a
                        If the functions intersect in ða; bÞ, then the integral yields an algebraic sum.  For example, if
                     gðxÞ¼ sin x and f ðxÞ¼ 0 then:
                                                    ð 2              2

                                                       sin xdx ¼ cos x    ¼ 0

                                                     0              0

                     VOLUMES OF REVOLUTION
                     Disk Method
                        Assume that f is continuous on a closed interval a @ x @ b and that f ðxÞ A 0.  Then the solid
                     realized through the revolution of a plane region R (bound by f ðxÞ, the x-axis, and x ¼ a and x ¼ b)
                     about the x-axis has the volume
                                                            ð b
                                                                  2
                                                       V ¼    ½ f ðxފ dx
                                                             a
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