# Practice Exam

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Practice Exam

January 25, 2009

π
1 Estimate the area under the graph f (x) = cos x from 0 to                                                    2   using four approximating rectangles and
midpoints. Sketch the graph and the rectangles.

Solution :
π π     π 3π    π 5π    π 7π    π     π   π     3π  π     5π  π     7π
f ( ) + f ( ) + f ( ) + f ( ) = cos ( ) + cos ( ) + cos ( ) + cos ( )
8 16    8 16    8 16    8 16    8     16  8     16  8     16  8     16

2 Express the following limit as a deﬁnite integral in the interval [2, 6]:
n
lim            xi ln(1 + x2 ) ∆x
i
n→∞
i=1

Solution :
n                                                6
lim              xi ln(1 + x2 ) ∆x =
i                                    x ln (1 + x2 ) dx
n→∞                                                      2
i=1

3 Evaluate the following integral:
10
x − 5 dx
0

Solution :
10                    5                        10
x2     5            x2        10           25        25
x−5 dx =             (5−x) dx+                 (x−5) dx = 5x−                               +           −5x         = 25−      +50−50− +25 = 25
0                     0                        5                                          2      0            2         5            2          2
Note you can also solve it by drawing the function and interpreting the area as the sum of the areas of two
triangles each with base 5 and height 5.

4 Find the derivative of the following function:
x2
1
g(x) =                     √          dt
tan x       2 + t4

Solution :
x2                             tan x
1                                 1                               2x        sec2 x
g(x) =                √          dt −                   √          dt , therefore g (x) = √        −√
0            2 + t4            0               2 + t4                          2 + x8    2 + tan4 x

1
√
x                                t2       1+u4
5 If F (x) =      1
f (t) dt, where f (t) =      1         u   du,        ﬁnd F (2).

Solution :
√                          √
1 + (x2 )4      1 + x8                     1 + 28  √
F (x) = f (x) , therefore F (x) = f (x) = 2x                              2
=2          , hence F (2) = 2          = 257
x               x                          2

4
6 If f (1) = 12, f is continuous and              1
f (x) dx = 17, what is the value of f (4).

Solution :
4
17 =           f (x) dx = f (4) − f (1) = f (4) − 12 , therefore f (4) = 17 + 12 = 29
1

7 Evaluate the following integral:
ex
dx
ex + 1

Solution :
ex              du
Let u = ex + 1 then du = ex dx hence                                      dx =          = ln u + C = ln(ex + 1) + C
ex + 1            u

8 Evaluate the following integral:
π
sin5 x dx
−π

Solution :
π
Since sin x is an odd function, then sin5 x is an odd function and therefore                           −π
sin5 x dx = 0

9 Evaluate the following integral:
10
(3x2 + x − 10) dx
0

Solution :
10
x3   x2                      10             100
(3x2 + x − 10) dx = 3            +    − 10x                     = 103 +       − 100 = 1000 + 50 − 100 = 950
0                                      3    2                       0               2

10 Who is the best superhero? (Circle the answer or write in another one)

Spider − M an , Superman , Batman , Captain America , Daredevil

Green Lantern , W olverine , Buf f y , Angel , T im Gunn
Ozymandias , Rorschach , M r. F antastic , El Santo , Other

Solution :
Gauss

2

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