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					                                               MATH 152, SPRING 2012
                                        COMMON EXAM II - VERSION B - SOLUTIONS

 Last Name:                                                                                  First Name:

 Signature:                                                                                Section No:

                                                        PART I: Multiple Choice (4 pts each)

1. The sequence whose terms are a n                                               n2 1
                                                                                    n2
     a.        increases and converges to 1.                                      Correct Choice
     b.        decreases and converges to 1.
     c.        increases and converges to 0.
     d.        decreases and converges to 0.
     e.        diverges.
                    2
               lim n 2 1
     Solution: n                                        lim 1            1                1 an             1        1 is less than 1 but getting bigger.
                     n                                  n                n2                                         n2
                                                                                            3
2. By substituting x                               3 tan , the integral                          x2 x2             9 dx becomes
                                                                                            0
                     /4
      a.                  81 tan 2 sec 2 d
                 0
                     /4
     b.                   81 tan 3 sec 2 d
                 0
                     /4
      c.                  27 tan 2 sec d
                 0
                     /4
     d.                   81 tan 2 sec 3 d                     Correct Choice
                 0
                 3
      e.             27 tan 2 sec 3 d
                 0

     Solution: x 2                      9 tan 2             x2       9                 9 tan 2         9           3 sec     dx         3 sec 2 d
          /4                                                                 /4
               9 tan 2             3 sec           3 sec 2 d                      81 tan 2 sec 3 d
      0                                                                  0

                          n
                  1                2n
3.                            n
                          6
     n 0

      a.         5
                14
     b.         27
                10
      c.         3
                10
     d.         33                 Correct Choice
                14
      e.        1
                3
                                               n
                                           1           2n                 1        n
                                                                                                  2n                1           1            6      3   33
     Solution:                                     n
                                               6                         6n                       6n           1        1   1       1        7      2   14
                                  n 0                          n 0                        n 0
                                                                                                                        6           3

                                                                                                                                                             1
4. Which of the following series diverges by the Test for Divergence?

         a.                sin                 1                Correct Choice
                                   2           n
              n 1

         b.                 ln n
                             n
              n 1

         c.                sin 1
                               n
              n 1

         d.                 n
                            n!
              n 1
         e. The Test for Divergence is inconclusive for all of the above series.
     Solution:
               l’H
     lim ln n
           n       lim 1
                       n                               0,       lim n       lim           1              0,        lim sin 1
                                                                                                                           n                   sin 0       0
     n             n                                            n   n!      n         n       1 !                  n


     lim sin                       1               sin                   1 So           sin                  1      diverges.
     n                      2      n                            2                               2            n
                                                                                  n 1

5. Find the length of the curve x                                        t2, y        t 3 , for 0            t         1.

         a. 2                13 13             1
            27
         b. 1
            27
         c. 1                13 13             1
            27
         d. 2                13 13             8
            27
         e. 1                13 13             8                Correct Choice
            27
                                           1                    2           2             1                                              1
                                                    dx                 dy                                2                  2
     Solution: L                                                                 dt                 2t             3t 2         dt           t 4   9t 2 dt
                                           0         dt                dt                 0                                              0
                                                            13                                      13
     u        4        9t 2            L           1                u du         1 2 u 3/2                       1 13 13                 8
                                                   18       4                    18 3               4            27
6. Find the surface area obtained by rotating the curve x                                                        cos 2t , y                  sin 2t , for
   0 t        , about the x-axis.
            4
         a.
              2
         b.
         c.
             4
         d. 4
         e. 2                    Correct Choice
     Solution: r                       y       sin 2t         because x-axis.
                      /4
                                       dx          2         dy 2          /4
                                                                                                                                     2                 2
     A                     2 r                                     dt         2 sin 2t                           2 sin 2t                2 cos 2t          dt
                  0                    dt                    dt          0
                       /4                                                       /4
                                                                  cos 2t
                            2 sin 2t 2 dt                   4                      2                         cos                     cos 0         2
                  0                                                 2         0                                         2

 2
7. Find the sum of the geometric series S                                4           8               16                .
                                                                         9           27              81

    a. S          2
    b. S          3
    c. S           4        Correct Choice
                   3
    d. S            4
                   15
    e. S           2
                   3
                                                                  4
   Solution: a              4        r        2        S          9                              4                4
                            9                 3                 1 2/3                        9       6            3

8. Which of the following statements is true regarding the improper integral                                                                           dx       ?
                                                                                                                                              1   ex        x

    a. The integral converges to 0.
    b. The integral converges because                                        dx                          dx           and                dx   converges.
                                                                1   ex               x               1   ex                          1   ex
                  Correct Choice
    c. The integral diverges because                                    dx                           dx           and                 dx diverges.
                                                            1   ex               x               1    x                          1      x
    d. The integral diverges because                                    dx                           dx           and                dx diverges.
                                                            1   ex            x                  1   ex                          1   ex
    e. The integral converges because                                        dx                              dx            and            dx converges.
                                                                1       ex           x               1        x                       1    x

   Solution:                1            1        and               1                    1               So (c) and (d) are wrong.
                    ex          x         x                ex            x               ex
         dx       converges and                   dx       diverges                  So (d) and (e) are wrong.
     1   ex                                   1    x
         1              0       So (a) is wrong.             (b) is correct by the Comparison Test.
    ex        x

9. The recursive sequence defined by a 1                                2, a n           1       5           4        converges. Find the limit.
                                                                                                             an

    a. 4           Correct Choice
    b. 5
    c. 1
    d.  5
        2
    e. 2
                                                                                                         2
              lim
    Solution: n a n              1   5           4              L            5       4               L        5L            4        0        L    4 L          1   0
                                              lim a n                                L
                                              n
    Limit must be 1 or 4.                a1        2, a 2           3, a 1                    11 , a n increasing from 2, Limit must be 4.
                                                                                              3



                                                                                                                                                                    3
10.            1  dx
           x2 x 1

      a. ln|x    1| 1 C
                     x
      b. ln|x| x  1 ln|x 1| C
      c. ln|x| 1 ln|x 1| C
                   x                                             Correct Choice
      d. ln|x 1| 1 C x
      e. ln|x 2 x 1 | C

      Solution:             1                     A         B                C         1     Ax x          1        Bx     1     Cx 2
                         x x 12                   x         x2           x       1
      x     0:             B   1                   x        1:                    C   1        Coeff of x 2 : 0 A C                            A   1
               1  dx                         1         1             1           dx        ln|x| 1 ln|x 1| C
           x2 x 1                            x         x2        x       1                        x

11. Compute                        dx .
                              1   1 x2

      a.
            4
      b.
          2
      c.  3               Correct Choice
           4
      d.
      e. 0

      Solution:                       dx         arctan x                                         3
                              1   1     x2                           1           2     4           4

12. Which of the following integrals gives the surface area obtained by rotating the curve
   y e 4x ,
    for 0 x 1, about the y-axis?
             1
                                        8x
      a.         2 x 1            16e        dx         Correct Choice
             0
             1
                         4x                  8x
      b.         2 e              1    16e        dx
             0
             e   4

      c.             2 y 1             1 dy
             1                        16y 2
             1
      d.                 16y 2        1 dy
             0       2
             1         ln y
      e.                    16y 2            1 dy
             0       8 y

      Solution: x-integral because y                             f x and 0             x     1.        r           x because y-axis.
            1           dy 2                                 1
                                                                                           4x 2
                                                                                                               1
                                                                                                                                     8x
      A       2 r 1            dx                                2 x 1                4e          dx               2 x 1       16e        dx
            0           dx                                   0                                                 0




 4
                                           e
13. The improper integral                        dx
                                           1    x ln x

     a.   diverges to    .
     b.   diverges to .    Correct Choice
     c.   converges to 1.
     d.   converges to 1.
     e.   converges to 1 1.
                        e
                                                                   e                     1
    Solution: u            ln x        du           dx                   dx                   du          ln 1       ln 0    0
                                                    x              1    x ln x           0
                                                                                              u



                                PART II: WORK OUT (48 points total)
     Directions: Present your solutions in the space provided. Show all your work neatly and
    concisely and box your final answer. You will be graded not merely on the final answer, but
    also on the quality and correctness of the work leading up to it.

14. (10 pts) Integrate                9        16x 2 dx.

    Solution: 4x            3 sin              4 dx        3 cos d                   9        16x 2              9     9 sin 2         3 cos

          9     16x 2 dx           3 cos 3 cos d                        9     1      cos 2 d                 9              sin 2         C
                                         4                              4            2                       8                2

    Draw a triangle or:                sin            4x           cos               1        sin 2              1      16x 2
                                                      3                                                                  9

          arcsin 4x                sin 2         sin cos                    4x 1             16x 2
                 3                   2                                      3                 9

          9     16x 2 dx       9 arcsin 4x                 4x 1              16x 2              C
                               8        3                  3                  9

15. (8 pts) Find the sum of the series: S                                    cos n            cos
                                                                       n 1                            n     1

                               k
    Solution: S k                   cos n            cos
                            n 1                            n   1

          cos       cos                cos               cos                             cos              cos                    cos       cos
                           2                    2              3                                k                k     1                         k   1
    S     lim S k     lim cos                  cos                      cos          cos 0            1          1      2
          k           k                              k     1




                                                                                                                                                         5
16. (10 pts) Integrate                            4x 2 1 dx.
                                                 2
                                                 x 1 x 2

     Solution:                4x 2 1                             Ax       B                C                4x 2         1         Ax        B x             2        C x2         1
                                 2
                             x 1 x 2                             x2       1            x       2

     4x 2       1        Ax           B x        2           C x2         1            A           C x2             B        2A x           C        2B
     A      C        4           B     2A            0           C       2B            1
     C      4       A            B      2A               4       A       4A            1           5A           5        A         1         B           2            C       3
             4x 2 1 dx                                x 2                     3        dx                   x                 2                  3        dx
            2
            x 1 x 2                                   x2 1                x        2                   x2       1        x2        1         x       2
         1 ln x 2            1        2 arctan x             3 ln|x           2|       C
         2

17. If the n-th partial sum of the series                                          a n is given by s n                             n        2,
                                                                                                                                       2n
                                                                          n 1


     (i) (5 pts) Find a 10 .

     Solution: a 10                   s 10       s9          10 2                  9 2                 6            11             1
                                                             2 10                  2 9                 10           18             90

     (ii) (5 pts) Find the sum of the series S                                                 an.
                                                                                        n 1



     Solution:                   an      lim s n             lim 1                 1
                                                                                   n
                                                                                                   1
                         n 1
                                         n                   n   2                                 2

18. (10 pts) Find the surface area obtained by rotating the curve y                                                                    x2        1 ln x, for 1                 x       2,
                                                                                                                                       4         2
   about the y-axis.

     Solution: x-integral because y                                       fx           and 1                x       2.         r        x because y-axis.
                2                                2                   2                                              2               2
                                        dy                                                     x        1                                                        x2       1         1
     A              2 r 1                             dx                 2 x 1                                          dx              2 x 1
                1                       dx                           1                         2        2x                         1                             4        2        4x 2
                2       2                                                     2                                                    2                                                    2
                  2 x x                      1         1 dx                     2 x x                   1           dx                  x2       1 dx                     x3       x
                1     4                      2        4x 2                    1     2                   2x                         1                                      3             1

                8        2              1        1               10
                3                       3                        3




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