Answer
$b=\frac{3}{2}+\frac{\sqrt 5}{2}=2.618, \frac{3}{2}−\frac{\sqrt 5}{2}=0.382$
Work Step by Step
We want to find the value of $b$ where the average value of $f(x)=2+6x-3x^{2}$ on the interval $[0,3]$ is equal to $3$.
$f(c)=3$
$3=\frac{1}{b-0}\int_{0}^{b}(2+6x-3x^{2})dx$
$3=\frac{1}{b}[2x+3x^{2}-x^{3}]_{0}^{b}$
Plug $b$ into $x$
$3=\frac{2b+3b^{2}-b^{3}}{b}= 2+3b-b^{2}$
$-b^{2}+3b-1=0$
Use quadratic formula to solve for b
$\frac{-3}{-2}±\frac{\sqrt{ 9-(4*-1*-1)}}{-2}=\frac{-3}{-2}±\frac{\sqrt{ 9-4}}{-2}=\frac{3±\sqrt 5}{2}$
$b=0.382, b=2.618$
Plug both values of $b$ into the main integral to verify
$3=\frac{1}{0.382}\int_{0}^{b}(2+6x-3x^{2})dx$
$3=\frac{1}{2.618}\int_{0}^{b}(2+6x-3x^{2})dx$
