# 2B

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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
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

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

6

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