let r be the region bounded by the functions f(x)=-2x² + 8 and g(x)=6x + 9 as shown in the diagram below…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $-2x^{2}+8 = 6x + 9$. Rearrange to $2x^{2}+6x + 1=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b=6$, $c = 1$, we get $x=\frac{-6\pm\sqrt{36 - 8}}{4}=\frac{-6\pm\sqrt{28}}{4}=\frac{-6\pm2\sqrt{7}}{4}=\frac{-3\pm\sqrt{7}}{2}$. Let $x_1=\frac{-3-\sqrt{7}}{2}$ and $x_2=\frac{-3 + \sqrt{7}}{2}$.

Step2: Set up integral for area

The area $A=\int_{x_1}^{x_2}[( - 2x^{2}+8)-(6x + 9)]dx=\int_{x_1}^{x_2}(-2x^{2}-6x - 1)dx$.

Step3: Integrate

$\int(-2x^{2}-6x - 1)dx=-\frac{2}{3}x^{3}-3x^{2}-x+C$.

Step4: Evaluate definite - integral

$A=\left[-\frac{2}{3}x^{3}-3x^{2}-x\right]_{x_1}^{x_2}=-\frac{2}{3}(x_2^{3}-x_1^{3})-3(x_2^{2}-x_1^{2})-(x_2 - x_1)$. Using a calculator with $x_1=\frac{-3-\sqrt{7}}{2}\approx - 2.823$ and $x_2=\frac{-3+\sqrt{7}}{2}\approx - 0.177$, we find $A\approx1.296$.

Answer:

$1.296$