Verified answer

Recall that the average value of f(x) on the interval [a, b] is given by:

F(avg) = 1/(b - a) ∫ f(x) dx (from x=a to b).

So, the average value of f(x) = 3 + 10x - 6x^2 on the interval [0, b] is:

F(avg) = 1/(b - 0) ∫ (3 + 10x - 6x^2) dx (from x=0 to b)

= (1/b)[3x + 5x^2 - 2x^3 (evaluated from x=0 to b)]

= (1/b)[(3b + 5b^2 - 2b^3) - (0 + 0 - 0)]

= 3 + 5b - 2b^2.

Since we want the average value to be 4, we have:

4 = 3 + 5b - 2b^2 ==> 2b^2 - 5b + 1 = 0.

Solving this with the Quadratic Formula yields b = (5 ± √17)/4.

I hope this helps!

Integral of f(x) from 0 to b is 3b+5b^2-2b^3 / (b-0) = 4. Or with quadratic formula b = 2.28 and b .22

Hint:

the average value of f(x) on the interval [0,b] is (1/b) int of f(x)dx , from x=0 to x=b