f(x)= 4x-1 [0,10] Average value over the given interval??? please help me!?

Answer 1

The average value of a given function f(x) over any interval [a,b] is given by the formula:
[integral f(x) from a to b]/(b-a)

The antiderivative of 4x-1 is 2x^2- x +C. So the average value is:
[integral 4x-1 from 0 to10]/(10)=
[2(10)^2-10]/10=190/10=19

So 19 is the average value over the interval [0, 10]

Answer 2

The only way I can think of doing it is the long way.

4(0(-1=-1
4(1)-1=3
4(2)-1=7
4(3)-1=11
4(4)-1=15
4(5)-1=19
4(6)-1=23
4(7)-1=27
4(8)-1=31
4(9)-1=35
4(10)-1=39

wow, they're all odd numbers..

anyways add all those together and divide by eleven:

[-1+3+7+11+15+19+23+27+31+35+39]/11
=19 is the average.

There must be some better way!

Answer 3

Your endpoints are (0,-1) and (10,39). Since the function is a straight line, the average is the average of the endpoints, (39 + (-1))/2 = 19 Notice that is also the value of f(5)