Calculus 1  Hour Exam 3   -   Math 220     Fall, 2011

 

 

 

 

 

 

1.

 

 

Suppose that  f  is an odd function which is integrable on the interval [-5,5].  If

 

      2                               3

      f(x) dx  =  4   and  ∫ f(x) dx = 10, then evaluate the following quantities.

0                                  2

 

            5                  3                        2                        2

a.       f(x) dx  +  ∫ f(x) dx         b. ∫ f(x) dx        c.  ∫ | f(x) | dx

0                           5                       -2                       -2

                                

  Answers:  a.  14          b.  0        c.  8

 

 

 

 

2.

 

                                                        18                         1

Evaluate the definite integrals.  a.   ∫ (1/2x) dx       b.  ∫ 8 / (1 + x2) dx

                                                        2                           0

Answers:  a.  ln 3        b. 

 

 

 

3.

 

 

 Evaluate the indefinite integrals.  a.   ∫ (12x / (1 + 3x2) dx       b.  ∫ tan x  sec5 x dx

 

Answers:  a.  2 ln(1 + 3x2) + C          b.  (1/5) sec5 x  +  C

                                                                       

 

 

4.

 

 

Evaluate the indefinite integral      ∫ 2x5 (x2 + 1)35 dx

 

Answer:  (1/38) (x2 + 1)38 – (2/27) (x2 + 1)37 + (1/36) (x2 + 1)36  +  C

 

 

 

 

 

 

 

 

5.

 

Let  R  be the finite region bounded by the graph of  f(x) = 5x - x2 and the x-axis

on the interval [0,5].  Set up, but do not evaluate, definite integrals which represent

the given quantities.  Use proper notation.

 

a.        The average value of  f  on the interval [0,5]

b.        The area of R

c.        The volume of the solid obtained by revolving R around the horizontal line y=-10.

d.       The volume of the solid obtained by revolving R around the vertical line x=8.

                             5                             5

Answers:  a. (1/5) ∫ (5x – x2) dx     b.  ∫ (5x – x2) dx

0                                   0

    5                                                                 5

c. ∫ π ( 5x – x2 + 10 ) 2 – π (10)2 ) dx         d.  ∫ 2π (8 – x) (5x – x2) dx

   0                                                                  0

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