Calculus 1  Hour Exam 3   -   Math 220     Fall, 2010

 

 

 

1.

 

 

The function  f(x)  =  10 x3 – 20 x  +  1  has one root in the interval  [1,2].  In order to

approximate this root, begin with an initial estimate of  x1  =  2 and use Newton’s

Method to obtain a second estimate  x2.  Write the value of  x2  in decimal form.

                                                                                                Answer:  x2  =  1.59

 

 

2.

 

Precisely state the Mean Value Theorem. 

 

 

 

3.

 

 

A function f(x)  has derivative  f ‘ (x)  =  6x2  + 5.  Find a formula for  f(x)  given that

its graph goes through the point  (1, 15).           Answer:    f(x)  =  2x3  + 5x  +  8

                                                                                

 

 

4.

 

 

Evaluate the following limit.  Be sure to use proper notation throughout your

evaluation of this limit.          n

                                      lim Σ  [ 14k/n2  -  4/n ]

                                          k = 1                                                          Answer:     3

 

 

 

 

5.

 

The height of a tree is currently 100 inches.  It is predicted that over the next 4 years

The tree’s height will increase by  10 - 3√t  inches per year where  t  represents the

number of years from now.  What will the tree’s height be 4 years from now?  Simplify

your answer.                                Answer:   124 inches

                                                                                                         

 

 

 

 

 

 

 

 

6.

 

 

 

Evaluate the following definite and indefinite integrals.

                                                                                     π

a.   I  =  ∫ ( 8/x  +  4 csc2 x  +  3 ) dx                   b.   I  =  ∫ ( 10 + 3 cos x ) dx

                                                                                   π/2

             2

c.  I  =  ∫ (6x  +  2 e-x ) dx                                   d.  I  =  ∫ x3 ( x4 + 7 )5  dx

            0

 

e.  I  =  ∫ sin3 x  cos5 x  dx                                  f.  I  =  ∫ ( 5 – 3 tan2 x)  dx

 

Answers:  a. I = 8 ln x  -  400 cot x  +  3x  +  C           b.  I  =  5π – 3

 

c.  I  =  14 – 2 e-2    d.  I  =  (1/24) (x4 + 7) + C    

 

e.  I  =  (1/4) sin4 x  - (1/3) sin6 x  +  (1/8) sin8 x  +  C       f.  I  =  8x – 3 tan x  +  C

 

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