the shaded region shown below is bounded by the functions f(x)=-2x² + x + 9 and g(x)=-x + 8 and the x…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $-2x^{2}+x + 9=-x + 8$. Rearrange to $2x^{2}-2x - 1=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b=-2$, $c=-1$, we get $x=\frac{2\pm\sqrt{4 + 8}}{4}=\frac{2\pm2\sqrt{3}}{4}=\frac{1\pm\sqrt{3}}{2}$. Also, find where $g(x)=-x + 8$ intersects the $x -$axis, set $g(x)=0$, then $x = 8$. And find where $f(x)=-2x^{2}+x + 9$ intersects the $x -$axis, set $f(x)=0$, $2x^{2}-x - 9=0$, $x=\frac{1\pm\sqrt{1+72}}{4}=\frac{1\pm\sqrt{73}}{4}$. From the graph, we need to integrate from the intersection point of $f(x)$ and $g(x)$ to $x = 8$. The intersection point of $f(x)$ and $g(x)$ in the relevant domain is $x=\frac{1+\sqrt{3}}{2}$.

Step2: Set up the integral for the area

The area $A=\int_{\frac{1 + \sqrt{3}}{2}}^{8}(-x + 8)dx$. The antiderivative of $-x+8$ is $-\frac{1}{2}x^{2}+8x$.

Step3: Evaluate the definite - integral

$A=\left(-\frac{1}{2}x^{2}+8x\right)\big|_{\frac{1+\sqrt{3}}{2}}^{8}=(-\frac{1}{2}(8)^{2}+8\times8)-\left(-\frac{1}{2}(\frac{1 + \sqrt{3}}{2})^{2}+8\times\frac{1+\sqrt{3}}{2}\right)$. First, $-\frac{1}{2}(8)^{2}+8\times8=-32 + 64=32$. Second, $-\frac{1}{2}(\frac{1+\sqrt{3}}{2})^{2}+8\times\frac{1+\sqrt{3}}{2}=-\frac{1}{2}\times\frac{1 + 2\sqrt{3}+3}{4}+4(1+\sqrt{3})=-\frac{4 + 2\sqrt{3}}{8}+4 + 4\sqrt{3}=-\frac{1}{2}-\frac{\sqrt{3}}{4}+4 + 4\sqrt{3}=\frac{7}{2}+\frac{15\sqrt{3}}{4}$. $A = 32-\left(\frac{7}{2}+\frac{15\sqrt{3}}{4}\right)=32-\frac{7}{2}-\frac{15\sqrt{3}}{4}=\frac{64 - 14}{2}-\frac{15\sqrt{3}}{4}=\frac{50}{2}-\frac{15\sqrt{3}}{4}=25-\frac{15\sqrt{3}}{4}\approx25 - 6.495=18.505$.

Answer:

$18.505$