the average value of the function f(x)=(x-1)^2 oon the interval from x=1 to x=5 is...?

Question

help would be appreciated

Answer 1

The average value of the function is given by the formula

A = [1/(b - a)] * Integral (a to b, f(x) ) dx

All we pretty much have to do is plug in the values.

A = [1/(5 - 1)] * Integral (1 to 5, (x - 1)^2)dx

OR

A = [1/4] * Integral (1 to 5, (x - 1)^2) dx

Taking the integral of (x - 1)^2 is easy, since the inside of the function is linear. All we have to do is use the reverse power rule; (1/3) (x - 1)^3

A = [1/4] * [ (1/3) (x - 1)^3 ] {evaluated from 1 to 5}

Pulling out the constant 1/3 (which we can do as a valid step even after the integral,

A = (1/12) * [x - 1]^3 {evaluated from 1 to 5}
A = (1/12) * [ (5 - 1)^3 - (1 - 1)^3 ]
A = (1/12) * [4^3 - 0^3]
A = (1/12) * [64]
A = 64/12

which, in reduced form, is equal to

A = 16/3

Answer 2

Definition: the average value of f(x) on the interval [a,b] is 1/(b-a) * integral of f(x) from x=a to x=b. Use the fundamental theorem: an anitderivative is (x_1)^3/3 evalaute at 1 and 5 and subtract to get (5-1)^3/3-(1-1)^3/3 = 64/3 then divide this by length of interval =5-1 = 4 to get average value is 16/3

Answer 3

f(1)=0
f(2)=1
f(3)=4
f(4)=9
f(5)=16
Avg=(0+1+4+9+16)/5=6