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Question 6

Page history last edited by PBworks 17 years ago

Question 6

 

You MAY NOT use a calculator to solve this question.

 

Let on the closed interval [0, 4π].

 

(a) Approximate F(2π) using four inscribed rectangles.

 

(b) Find F'(2π).

 

(c) Find the average value of F'(x) on the interval [0, 4π].

 

Solution

 

(a) This question asks us to approximate the value of F(x) from 0 to 2π. To do this, we may use the methods we learned in class. The easiest approximation we can use (note that we can't use a calculator) that suits our need is the midpoint sum. Here's a graph of what we want to do:

 

 

 

 

(No one edited my work, so I did... is that allowed Mr. K?)

 

(b) Finding F'(2π).

 

Note: S = The indefinite integral sign. 0 S 2π reads "integrate from zero to two pi".

 

F(x)            = (0 S x) [cos(t/2) + (3/2)] dt

F(x)(dt/dx) = (0 S x) [cos(t/2) + (3/2)] (dt/dx)           The "0 S x" and (dt/dx) cancel each other out.

 

F'(x)           = cos(x/2) + (3/2)                                  Write what's left.

F'(2π)         = cos(2π/2) + (3/2)                                Plug in "2π" where x is.   

 

                  = cos(π) + (3/2)                                     Solve.

                  = -1 + (3/2)  

                  = (-2/2) + (3/2)

                  = 1/2

 

(c) Finding the average value of F'(x) on the interval [0, 4π]

 

Average value =    1      (a S b) F'(x) dx

                           (b-a)

 

                        =     1    (0 S 4π) cos(x/2) + (3/2)

                           (4π-0)

                        =    1   [sin(x/2)(1/2) + (3/2)x](from zero to four)

                             4π

                        =    1   [(0+6π) - (0)]  

                             4π

                        =    1   (6π)

                             4π

                        = 

                            4π

                        =  3

                            2

 

 

 

 

 

 

 

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