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Find the numbers b such that the average value of f(x) = 15x2 − 42x + 6 on the interval [0, b] is equal to 2.

User Cikenerd
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1 Answer

3 votes
The average value (a) of f(x) on the interval [0, b] is given by

a= (1)/(b) \int\limits^b_0 {(15x^(2)-42x+6)} \, dx = 5b^(2) -21b +6
You want this value equal to 2, so you have

5b^(2) -21b +6 = 2

(5b-1)(b-4)=0

The values of b are 0.2 and 4.
User NeoNexus DeMortis
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