calculus

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Find the average value of the function
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 -

    Let 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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