# What is the average value of the function # f(x)=(x-1)^2# on the interval #[1,5]#?

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To find the average value of the function ( f(x) = (x - 1)^2 ) on the interval ([1, 5]), you need to calculate the definite integral of the function over that interval and then divide it by the length of the interval.

[ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) , dx ]

Substituting the values (a = 1) and (b = 5) into the formula:

[ \text{Average value} = \frac{1}{5 - 1} \int_{1}^{5} (x - 1)^2 , dx ]

Now, calculate the definite integral:

[ \int_{1}^{5} (x - 1)^2 , dx = \left[ \frac{(x - 1)^3}{3} \right]_{1}^{5} = \frac{(5 - 1)^3}{3} - \frac{(1 - 1)^3}{3} = \frac{64}{3} ]

Substituting this value into the formula for the average value:

[ \text{Average value} = \frac{1}{5 - 1} \times \frac{64}{3} = \frac{64}{12} = \frac{16}{3} ]

So, the average value of the function ( f(x) = (x - 1)^2 ) on the interval ([1, 5]) is ( \frac{16}{3} ).

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