question the shaded region shown below is bounded by the functions f(x)=-x² + 9 and g(x)=-1.25x + 7, the y…

Explanation:

Step1: Set up the integral for area

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, $f(x)=-x^{2}+9$, $g(x)=-1.25x + 7$, $a = 0$, $b = 3$. Since $f(x)\geq g(x)$ on the interval $[0,3]$, the area formula becomes $A=\int_{0}^{3}[(-x^{2}+9)-(-1.25x + 7)]dx=\int_{0}^{3}(-x^{2}+1.25x + 2)dx$.

Step2: Integrate term - by - term

Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int(-x^{2}+1.25x + 2)dx=-\frac{x^{3}}{3}+\frac{1.25x^{2}}{2}+2x+C$.

Step3: Evaluate the definite integral

$A=\left[-\frac{x^{3}}{3}+\frac{1.25x^{2}}{2}+2x\right]_{0}^{3}$. First, substitute $x = 3$: $-\frac{3^{3}}{3}+\frac{1.25\times3^{2}}{2}+2\times3=-9+\frac{1.25\times9}{2}+6$. Then, $-9+\frac{11.25}{2}+6=-9 + 5.625+6=2.625$. Substitute $x = 0$ which gives $0$. So $A=2.625$.

Answer:

$2.625$