MHB Why Does Integration Matter in Solving for the Average Value of a Function?

[nghijen]

I have attempted to solve the question but I still do not understand. Can someone please help me?

 

[skeeter]

average value of a function involves a definite integral over an interval ... from your shading, is the given interval $[0,u]$ ?

 

[nghijen]

average value of a function involves a definite integral over an interval ... from your shading, is the given interval $[0,u]$ ?

I am assuming it is [0, u] !

 

[skeeter]

I am assuming it is [0, u] !

$\displaystyle 0 = \dfrac{1}{u-0} \int_0^u x^2(5-x) \, dx$

work it ...

 

[HOI]

I frankly can't understand what you think you are doing! Much of what you are doing is determining where f(x)= 0. Okay that is at x= 5, which you have, and x= 0, which you ignore, but that is irrelevant anyway! The average value of a function, f(x), over interval a to b is the integral of f, from a to b, divided by b- a. But you show no attempt at integration!

Oh, and $-5^2(5- 5)= 0$, not -25! That is why you got x= 5 as a solution to $-x^2(x- 5)= 0$!