Problem

Find the average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ The average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ is (Simplify your answer.)

Solution

Step 1 :The average value of a function \(f(t)\) on the interval \([a, b]\) is given by the formula: \[\frac{1}{b-a}\int_{a}^{b}f(t)dt\]

Step 2 :In this case, \(f(t) = (t-5)^2\), \(a = 0\), and \(b = 12\). So we need to compute the integral of \((t-5)^2\) from \(0\) to \(12\) and then divide by \(12 - 0 = 12\).

Step 3 :The average value of the function is calculated to be 13.

Step 4 :Final Answer: The average value of the function \(f(t)=(t-5)^{2}\) on \([0,12]\) is \(\boxed{13}\).

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Source: https://solvelyapp.com/problems/8004/

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