**Question:**

Find the total area of the region shaded in the following diagram:

$\displaystyle y = 3x^2 - 12x + 9$

**Attempt:**

$\displaystyle y = 3x^2 - 12x + 9$

$\displaystyle a = 3, b = -12, c = 9$

$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle x = \frac{-(-12) \pm \sqrt{(-12)^2-4\times3\times9}}{2\times3}$

$\displaystyle x = \frac{12 \pm \sqrt{36}}{6}$

$\displaystyle x = \frac{12 + 6}{6}$ , $\displaystyle x = \frac{12 - 6}{6}$

$\displaystyle x = 3$ , $\displaystyle x = 1$

$\displaystyle =3x^2-12x+9$

$\displaystyle =\frac{3x^{2+1}}{2+1} - \frac{12x^{1+1}}{1+1} + 9x$

$\displaystyle =\frac{3x^{3}}{3} - \frac{12x^{2}}{2} + 9x$

$\displaystyle =x^3-6x^2+9x$

$\displaystyle = [x^3-6x^2+9x]^3_1$

$\displaystyle = [(3)^3-6(3)^2+9(3)] - [(1)^3-6(1)^2+9(1)]$

$\displaystyle =-4$