Question

How do you find average value of `f(x) = sinx` over `[0, pi/2]` and verify the Mean Thereom for `f(x)=x^2` over [0,1]?

Answer 1

The "average value" of a function has to do with integration. This question is posted under "Average Rate of Change Over an Interval" I think that you meant to ask:

How do you find average rate of change of `f(x) = sinx` over `[0, pi/2]` and verify the Mean Theorem for `f(x)=x^2` over [0,1]?

(If that wasn't the question, i am sorry for my misunderstanding.)

The average rate of change of function `f` over interval `[a,b}` is:

`(f(b)-f(a))/(x-a)`

(Yes, that is also a difference quotient and also the slope of a secant line.)

For this function on this interval you should get:
`(sin(pi/2)-sin(0))/(pi/2 - 0) = 1/(pi/2) = 2/pi`
is

The average value of `sinx` over `[0,pi/2]` is
`1/(pi/2-0) int_0^(pi/2) sinx dx = 2/pi [-cosx]_0^(pi/2)`

`= 2/pi[1] = 2/pi` (again)

For the second question, I think what your teacher/textbook wants you to do is the following:

The function `f(x) = sinx` on the interval `[0, pi/2]` satisfies the hypotheses because
`sinx` is continuous everywhere, so it is certainly continuous on `[0, pi/2]`.

`sinx` is differentiable on the interval `(0, pi/2)` as required, because the derivative is `cosx` which exists for every `x` in the interval `(0, pi/2)`. (and, in fact for every real `x`).

The conclusion of the theorem says : there is a number `c` in the interval `(0, pi/2)` with `f'(c) = (sin(pi/2)-sin(0))/(pi/2 - 0)` .

(The instantaneous rate of change at some `c` inside the interval is equal to the average rate of change over the interval.)

You might try to find the `c` to "verify" that it exists.
Since `f'(x)=cosx`, you need to solve:

`cosx = 2/pi`.

OK! We're not going to be able to solve this, but we can use the Intermediate Value Theorem to show that there is a solution.

Since `pi ~~ 3.14`, we must have `2/pi <1` And clearly `2/pi > 0`.

`cosx` is continuous on `[0, pi/2]`,
and `cos0 = 1> 2/pi` and also `cos(pi/2) = 0 < 2/pi`,

so `2/pi` is between `cos0` and `cos (pi/2)`,

so the Intermediate Value Theorem tells us that there is a `c` in `(0, pi/2)` with `cosc= 2/ pi`