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.)

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

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

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 $13$ .

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.)

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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.)