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

  • MHB
  • Thread starter nghijen
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In summary, the average value of a function involves a definite integral over an interval and is calculated by dividing the integral of the function over the interval by the length of the interval. The given interval in this problem is assumed to be [0, u] and the solution involves solving the definite integral. However, the approach taken by the person in the conversation is unclear and does not consider the necessary steps for integration.
  • #1
nghijen
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0
I have attempted to solve the question but I still do not understand. Can someone please help me?
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  • #2
average value of a function involves a definite integral over an interval ... from your shading, is the given interval $[0,u]$ ?
 
  • #3
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] !
 
  • #4
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 ...
 
  • #5
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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