the shaded region shown below is bounded by the functions f(x)=-x² - 2x + 8 and g(x)=2.25x + 2, the y - axis…

Explanation:

Step1: Determine the upper - lower functions

On the interval $[0,2]$, we need to find which function is on top and which is on the bottom. Let's find the difference $h(x)=f(x)-g(x)=-x^{2}-2x + 8-(2.25x + 2)=-x^{2}-4.25x + 6$. We can also test a value in the interval $[0,2]$, say $x = 1$. $f(1)=-1^{2}-2\times1 + 8=5$ and $g(1)=2.25\times1+2 = 4.25$. So $f(x)\geq g(x)$ on $[0,2]$.

Step2: Use the area formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $a = 0$, $b = 2$, $f(x)=-x^{2}-2x + 8$ and $g(x)=2.25x + 2$. So $A=\int_{0}^{2}(-x^{2}-2x + 8-(2.25x + 2))dx=\int_{0}^{2}(-x^{2}-4.25x + 6)dx$.

Step3: Integrate term - by - term

We know that $\int(-x^{2}-4.25x + 6)dx=-\frac{1}{3}x^{3}-\frac{4.25}{2}x^{2}+6x+C$.

Step4: Evaluate the definite integral

$A=\left[-\frac{1}{3}x^{3}-\frac{4.25}{2}x^{2}+6x\right]_{0}^{2}=-\frac{1}{3}(2)^{3}-\frac{4.25}{2}(2)^{2}+6\times2-\left(-\frac{1}{3}(0)^{3}-\frac{4.25}{2}(0)^{2}+6\times0\right)$. $A=-\frac{8}{3}-8.5 + 12$. $A=-\frac{8}{3}+3.5$. $A=-\frac{8}{3}+\frac{7}{2}=\frac{-16 + 21}{6}=\frac{5}{6}\approx0.833$.

Answer:

$0.833$