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

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SUMMARY

The discussion centers on the calculation of the average value of a function using definite integrals, specifically the function \( f(x) = x^2(5-x) \) over the interval \([0, u]\). The average value is defined as \(\frac{1}{u-0} \int_0^u f(x) \, dx\). Participants highlight the importance of correctly identifying the interval and performing the integration, noting that the average value formula requires proper execution of the integral. Miscalculations, such as incorrectly evaluating \( -5^2(5-5) \), lead to confusion regarding the function's roots.

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nghijen
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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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