Average Value of a Function

Average value = $\displaystyle \large \frac{1}{b - a} \int_{a}^{b} f(x) dx$

$\displaystyle \large 3 = \frac{1}{b - 0} \int_{0}^{b} (-3x^2 + 6x + 2) \ dx$

$\displaystyle \large 3b = \int_{0}^{b} (-3x^2 + 6x + 2) \ dx$

$\displaystyle \large 3b = -x^3 + 3x^2 + 2x | ^b_0$

$\displaystyle \large 3b = -b^3 + 3b^2 + 2b$

$\displaystyle \large b^3 - 3b^2 + b = 0$

$\displaystyle \large b(b^2 - 3b + 1) = 0$

$\displaystyle \large b \neq 0$ because $\displaystyle \large \frac{1}{0 - 0} \int_{0}^{0} f(x) dx$ is undefined

$\displaystyle \large b^2 - 3b + 1 = 0$

Use the quadratic formula:

$\displaystyle \large b = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{3^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$