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Posted by **li** on Saturday, December 3, 2011 at 4:06pm.

f(x)=8x^2-5x+6 , on the interval [3,5]. Find the value of x-coordinate at which the function assumes it's average value.

what is the average value = to ?

what is the x coordinate = to ?

Thanks

- calculus -
**MathMate**, Sunday, December 4, 2011 at 1:02pmLet f(x)=8x^2-5x+6

The average value is the definite integral divided by the interval.

I=∫f(x)dx

=∫(8x^2-5x+6)dx

=[(8/3)x^3-(5/2)x^2+6x] from 3 to 5

=700/3

Average value, a

= I/(5-3)

= 350/3

To find x where f(x)=a, solve for x in

f(x) = 8x^2-5x+6 = 350/3

However, out of the two solutions, x=-3.42 and x=4.045, you will report the solution which is within the interval [3,5] and reject all solutions outside of the interval.

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