Question

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

Answer 1

The average value (a) of f(x) on the interval [0, b] is given by
$a= \frac{1}{b} \int\limits^b_0 {(15x^{2}-42x+6)} \, dx = 5b^{2} -21b +6$
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.