Page 180 - Calculus Workbook For Dummies
P. 180

164       Part IV: Integration and Infinite Series



                     9. Simplify as in Step 6.                        Done! Finally! That’s the formula for
                                                                      approximating the area under
                                       2
                                5  n  5   5  n  5                             2
                              = !c  n  i + ! cm  n  3  n  im          f x =  x +  3 x from 0 to 5 with n
                                                                       ^ h
                                n
                                 i 1       i 1
                                  =
                                                                      rectangles — the more you use, the
                                            =
                                           n
                                  n
                                         5
                                5
                              = !  25 2  i + ! 15  i                  better your estimate. I bet you can’t
                                       2
                                         n
                                             n
                                n
                                           =
                                  =
                                 i 1  n   i 1                         wait to do one of these problems on
                               125  n  2  75  n                       your own.
                              =  3  ! i +  2  i !
                                n  i 1=  n  i 1=
                                                                      Check this result by plugging 20 into n to
                   10. Now replace the sigma sums with the            see whether you get the same answer as
                       expressions for the sums of integers           with the 20-rectangle version of this
                       and squares of integers like you did in        problem.
                       Step 7.                                                2
                                                                        475 20 +  600 20 +  125
                                                                                       h
                                                                                    ^
                                                                           ^
                                                                              h
                              ^
                                                  ^
                        125  n n +  1 2 ^h  n +  1h  75  n n +  1h    =             2        .  84 .2
                                                                                 ^
                       =  3 e             o  +  2 e     o                      6 20h
                         n         6         n      2
                                                                      It checks.
                                             2
                            2
                        250 n +  375 n +  125  75 n +  75 n
                       =                +
                               6 n  2        2 n 2
                            2
                        475 n +  600 n +  125
                       =         2
                               6 n
                                                                             9
                             10
               5.   Evaluate ! 4.                              6.   Evaluate !^ - 1 ^h i  i +  1h .
                                                                                       2
                             =
                             i 1                                            i =  0
                Solve It                                        Solve It
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