# What is the average value of a function #f(x) x^2 - 2x + 5# on the interval [2,6]?

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To find the average value of a function ( f(x) = x^2 - 2x + 5 ) on the interval [2,6], we first need to find the definite integral of the function over that interval, and then divide the result by the length of the interval.

The definite integral of ( f(x) ) from 2 to 6 can be calculated as follows:

[ \int_{2}^{6} (x^2 - 2x + 5) , dx ]

Using the power rule of integration, we find:

[ \left[ \frac{x^3}{3} - x^2 + 5x \right]_{2}^{6} ]

Substituting the upper and lower limits:

[ \left( \frac{6^3}{3} - 6^2 + 5 \cdot 6 \right) - \left( \frac{2^3}{3} - 2^2 + 5 \cdot 2 \right) ]

Solving this, we get:

[ \left( \frac{216}{3} - 36 + 30 \right) - \left( \frac{8}{3} - 4 + 10 \right) ] [ \left( 72 - 36 + 30 \right) - \left( \frac{8}{3} - 4 + 10 \right) ] [ ( 66 ) - \left( \frac{8}{3} - 4 + 10 \right) ] [ ( 66 ) - \left( \frac{8}{3} + 6 \right) ] [ ( 66 ) - \left( \frac{8}{3} + \frac{18}{3} \right) ] [ ( 66 ) - \frac{26}{3} ] [ \frac{198}{3} - \frac{26}{3} = \frac{172}{3} ]

Finally, to find the average value, we divide this result by the length of the interval [2,6], which is ( 6 - 2 = 4 ):

[ \frac{172}{3} \div 4 = \frac{43}{3} \approx 14.33 ]

So, the average value of the function ( f(x) = x^2 - 2x + 5 ) on the interval [2,6] is approximately ( \frac{43}{3} ) or 14.33.

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