Better Students Ask More Questions.
Using the mean value theorem, obtain the mean value of the function f(x) = sin(x) in...
1 Answer | add yours
(Level 1) Associate Educator, Expert, Newton
To find the average value of `f(x)` on the interval [a,b], you compute:
`1/(b-a) int_a^b f(x) dx`
So in our case, we would do:
`1/(pi-0) int_0^pi "sin" x dx`
`=1/pi [-"cos" x]|_0^pi`
`=1/pi (1 + 1) = 2/pi`
So, the average value of `"sin" x` on `[o,pi]` is `2/pi`
The value of x where the average occurs:
We want to find c in `[0,pi]` such that `"sin" c = 2/pi`
Take `"Sin"^(-1) (2/pi) ~~.690107 `
This is different from the mean value theorem.
The mean value theorem states that:
If f(x) is continuous on [a,b] and differentiable on (a,b), then there is some c in (a,b) where:
`f'(c) = (f(b)-f(a))/(b-a)`
That is, there is some place where the derivative is equal to the slope of the secant line that connects the endpoints
For us, sin(pi) = sin(0) = 0
f'(x) = cos (x)
So the mean value theorem says:
There is some c in [0,pi] such that:
`"cos" (c) = (0-0)/(pi-0) = 0`
And sure enough, at c=`pi/2` , cos(c) = 0
Posted by mlehuzzah on December 12, 2012 at 6:13 PM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students.