the shaded region shown below is bounded by the functions f(x)=-x² + 9 and g(x)=0.5x + 6, the y - axis and…

Explanation:

Step1: Determine the upper - lower functions

The upper function is $f(x)=-x^{2}+9$ and the lower function is $g(x)=0.5x + 6$ on the interval $[0,3]$. The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. So, $A=\int_{0}^{3}[(-x^{2}+9)-(0.5x + 6)]dx=\int_{0}^{3}(-x^{2}-0.5x + 3)dx$.

Step2: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have: $\int(-x^{2}-0.5x + 3)dx=-\frac{x^{3}}{3}-0.5\times\frac{x^{2}}{2}+3x+C=-\frac{x^{3}}{3}-\frac{x^{2}}{4}+3x+C$.

Step3: Evaluate the definite integral

$A=\left[-\frac{x^{3}}{3}-\frac{x^{2}}{4}+3x\right]_{0}^{3}$. First, substitute $x = 3$: $-\frac{3^{3}}{3}-\frac{3^{2}}{4}+3\times3=-\frac{27}{3}-\frac{9}{4}+9=-9-\frac{9}{4}+9=-\frac{9}{4}$. Then substitute $x = 0$: $-\frac{0^{3}}{3}-\frac{0^{2}}{4}+3\times0 = 0$. $A=-\frac{9}{4}-0=- 2.250$ (but area is non - negative, so we take the absolute value).

Answer:

$2.250$