Question

What is the average value on the function `f(x)=4x^(2)-8x+5` on the interval [-1,2]?

Answer 1

I found `A_V=5`

Explanation:

Have a look:

Graphically:

Answer 2

5

Explanation:

The average value of a function over an interval is the height a rectangle on that interval would need to have, if we want the area of the rectangle to be equal to the area under the graph of the function.

To find that height, find the area under the graph, then divide by the length of the interval.

Average value = `(int_-1^2 (4x^(2)-8x+5) dx)/(2-(-1))`

This is probably better written: `1/(2-(-1)) int_-1^2 (4x^(2)-8x+5) dx`

`int_-1^2 (4x^(2)-8x+5) dx = (4x^3)/3 - 4x^2 +5x]_-1^2 =15`

So the average value of `f(x) = 4x^(2)-8x+5` on `[-1,2]` is `1/3 * 15 =5`