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# Question 6

last edited by 16 years, 5 months 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