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

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Integration is crucial for determining the average value of a function over a specified interval, typically calculated using a definite integral. The discussion centers on the average value formula, which involves integrating the function and dividing by the interval's length. There is confusion regarding the integration process and the specific interval being used, which is assumed to be [0, u]. Additionally, there are corrections needed in the calculations presented, particularly regarding the evaluation of the integral and the identification of critical points. Understanding these concepts is essential for accurately solving for the average value of a function.
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I have attempted to solve the question but I still do not understand. Can someone please help me?
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average value of a function involves a definite integral over an interval ... from your shading, is the given interval $[0,u]$ ?
 
skeeter said:
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] !
 
nghijen said:
I am assuming it is [0, u] !

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

work it ...
 
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 $-x^2(x- 5)= 0$!
 
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