the shaded region shown below is bounded by the functions f(x)=-2x² + 3x + 9 and g(x)=-x + 8 and the x and y…

Explanation:

Step1: Find intersection point

Set $f(x)=g(x)$, so $-2x^{2}+3x + 9=-x + 8$. Rearrange to $2x^{2}-4x - 1=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b=-4$, $c=-1$, we get $x=\frac{4\pm\sqrt{16+8}}{4}=\frac{4\pm\sqrt{24}}{4}=\frac{4\pm2\sqrt{6}}{4}=1\pm\frac{\sqrt{6}}{2}$. We take the positive - valued root $x = 1+\frac{\sqrt{6}}{2}\approx1 + 1.225=2.225$ since we are interested in the region in the first - quadrant. Also, find the $x$ - intercept of $f(x)$ by setting $f(x)=0$, $-2x^{2}+3x + 9 = 0$, or $2x^{2}-3x - 9=0$, factoring gives $(2x + 3)(x - 3)=0$, so $x = 3$ is the positive $x$ - intercept. The $x$ - intercept of $g(x)$ is $x = 8$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x = 0$ to $x=a$ is given by $A=\int_{0}^{a}|f(x)-g(x)|dx$. Here, for $0\leq x\leq1+\frac{\sqrt{6}}{2}$, $f(x)\geq g(x)$, and the area is $A=\int_{0}^{1+\frac{\sqrt{6}}{2}}((-2x^{2}+3x + 9)-(-x + 8))dx=\int_{0}^{1+\frac{\sqrt{6}}{2}}(-2x^{2}+4x + 1)dx$.

Step3: Evaluate the integral

Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(-2x^{2}+4x + 1)dx=-\frac{2}{3}x^{3}+2x^{2}+x+C$. Then $\left[-\frac{2}{3}x^{3}+2x^{2}+x\right]_{0}^{1+\frac{\sqrt{6}}{2}}=-\frac{2}{3}(1+\frac{\sqrt{6}}{2})^{3}+2(1+\frac{\sqrt{6}}{2})^{2}+(1+\frac{\sqrt{6}}{2})$. Using a calculator: $(1+\frac{\sqrt{6}}{2})^{2}=1+\sqrt{6}+\frac{6}{4}=2.5+\sqrt{6}\approx2.5 + 2.449=4.949$. $(1+\frac{\sqrt{6}}{2})^{3}=(1+\frac{\sqrt{6}}{2})(1+\frac{\sqrt{6}}{2})^{2}=(1+\frac{\sqrt{6}}{2})(2.5+\sqrt{6})=2.5+\sqrt{6}+1.25\sqrt{6}+3=5.5 + 2.25\sqrt{6}\approx5.5+2.25\times2.449=5.5 + 5.51=11.01$. $-\frac{2}{3}(1+\frac{\sqrt{6}}{2})^{3}+2(1+\frac{\sqrt{6}}{2})^{2}+(1+\frac{\sqrt{6}}{2})=-\frac{2}{3}\times11.01+2\times4.949+(2.225)$ $=-7.34+9.898 + 2.225=4.783$.

Answer:

$4.783$