Find the average value in a calculus approach

silverbell
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Homework Statement



Find the average value of : f(x) = 4sinx + 4cosx on the interval [0, 16pi/6]

Average value: ?

Homework Equations



Integrals

The Attempt at a Solution



1/ [(16pi/6) -0] ∫ from 0 to 16pi/6 4sinx + 4cosx dx

[3/8pi] ∫ from 0 to 16pi/6 4sinx + 4cosx dx <------simplify

[3/8pi] ∫ from 0 to 16pi/6 4sinx - 4cosx <------ integrated

[3/8pi] [4sin(16pi/6) - 4cos(16pi/6)] - [3/8pi] [ 4sin(0) - 4cos(0)] <----substitution

I get 12.296 but the answer isn't right. I don't know what I'm doing wrong. Please help me understand what I'm doing wrong. Thank you very much. :)
 
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What did you get for 3/(8π),? (Yes, the parentheses are important.)

    "    sin(16π/6) ?

    "    cos(16π/6) ?

    "    4sin(0) - 4cos(0) ?

I would estimate that the answer is considerably smaller than 4, maybe less than 1.
 
I got for

3/(8π) ---> 0.119

sin(16π/6) ---> 0.866

cos(16π/6) ---> -0.5

4sin(0) - 4cos(0) ---> 0 - 4
 
Looks like you integrated it properly. I think you evaluated the terms incorrectly.

I suggest you factor the expression as much as possible to remove the 4's. That cleans up your expression and reduces chance of error when cranking through the numbers.
 
LawrenceC said:
Looks like you integrated it properly. I think you evaluated the terms incorrectly.

I suggest you factor the expression as much as possible to remove the 4's. That cleans up your expression and reduces chance of error when cranking through the numbers.


Thanks for the tip. I got the answers. :D
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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