Why is the average value of function at the end point is equal to the average value?

Question

for eg: f(x) in the range (-3,3) the value of function at 3=[f(3)+f(-3)]/2.....how was this formula derived?please help me

Answer 1

What do you mean by an "average" value of a function? There is no such term.
The closest thing to it I can think of is the "mean value theorem". It says, that if you divide the difference between function values at the ends of an interval over the length of the interval, you will get a value for that function's derivative at some point inside the interval.
Is this what you are talking about?
If so, you can read its detailed explanation and a proof here:
http://marauder.millersville.edu/~bikenaga/calc/mvt/mvt.html

Answer 2

This question doesn't really make sense. The calculus definition of the average value of the function is: 1 / (b-a) * Integral (from a to b) of f(x)dx. That is, integrate the function over the domain you are interested in, and divide by the length of that domain. If you look at a simple function, say, f(x) = x from -1 to 1, the average value is 0, but the value of the function at the endpoints is -1 and 1 respectively. In other words, you can't really find the "average value" of a function at a point (it's just the value of the function at that point) and that is almost never equal to the average value of the function.

Answer 3

An average value is a constant value, You can't change it. All values of a function are the source for it.

So at the endpoint of a function also this average value is valid.

But if You want to be exact, You can take also the actual function-value of the endpoint of Your function. This value is valid only for the endpoint and not for the whole function.

Answer 4

This is not true. This is true for only odd functions and not even function for example

f(x) = x^2

at 3 9
at -3 9
at 2 and -2 4
at 1 and -1 1
at 0 0
average = (9+9+4+4+1+1+0)/7 = 4
average at end points 9
so this is not correct