The average value (a) of f(x) on the interval [0, b] is given by
[tex] a= \frac{1}{b} \int\limits^b_0 {(15x^{2}-42x+6)} \, dx = 5b^{2} -21b +6[/tex]
You want this value equal to 2, so you have
[tex] 5b^{2} -21b +6 = 2 [/tex]
[tex] (5b-1)(b-4)=0 [/tex]
The values of b are 0.2 and 4.
