Single Variable Calculus, Chapter 6, 6.5, Section 6.5, Problem 14

Determine the numbers $b$ such that the average value of $f(x) = 2 + 6x - 3x^2$ on the interval $[0, b]$ is eqaul to 3.


$
\begin{equation}
\begin{aligned}

& f_{ave} = \frac{1}{b - a} \int^b_a f(x) dx
\\
\\
& 3 = \frac{1}{b - 0} \int^b_0 (2 + 6x - 3x^2) dx
\\
\\
& 3b = \left[ 2x + \frac{6x^2}{2} - \frac{3x^3}{3} \right]^b_0
\\
\\
& 3b = \left] 2(b) + \frac{6 (b)^2}{2} - \frac{3 (b)^3}{3} \right] - \left[ 2(0) + \frac{6(0)^2}{2} - \frac{3(0)^2}{3} \right]
\\
\\
& 3b = 2b + 3b^2 - b^3
\\
\\
& b^3 - 3b^2 + b = 0
\\
\\
& \text{We have,}
\\
\\
& b = 0 \text{ and } b^2 - 3b + 1 = 0
\\
\\
& \text{By applying Quadratic Formula}
\\
\\
& b = 2.6180 \text{ and } b = 0.3820

\end{aligned}
\end{equation}
$


Therefore, the values of $b$ are..

$b = 0, b = 2.6180$ and $b = 0.3820$