question the shaded region shown below is bounded by the functions f(x)= -2x² + 9 and g(x)= -1.75x + 8 and…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $-2x^{2}+9=-1.75x + 8$. Rearrange to $2x^{2}-1.75x - 1=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b=-1.75$, $c=-1$, we get the intersection - points.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $f(x)-g(x)=-2x^{2}+9-(-1.75x + 8)=-2x^{2}+1.75x + 1$. We need to find the appropriate limits of integration from the intersection points and $x = 0$.

Step3: Evaluate integral

Use a calculator to evaluate the definite integral $\int_{0}^{x_{0}}(-2x^{2}+1.75x + 1)dx$, where $x_{0}$ is the positive root of $2x^{2}-1.75x - 1=0$. The positive root of $2x^{2}-1.75x - 1=0$ is $x=\frac{1.75+\sqrt{(-1.75)^{2}-4\times2\times(-1)}}{4}=\frac{1.75+\sqrt{3.0625 + 8}}{4}=\frac{1.75+\sqrt{11.0625}}{4}\approx\frac{1.75 + 3.326}{4}\approx1.27$. Then $\int_{0}^{1.27}(-2x^{2}+1.75x + 1)dx=\left[-2\times\frac{x^{3}}{3}+1.75\times\frac{x^{2}}{2}+x\right]_{0}^{1.27}$. [ \begin{align*} &-2\times\frac{(1.27)^{3}}{3}+1.75\times\frac{(1.27)^{2}}{2}+1.27-0\ =&-\frac{2\times2.048}{3}+\frac{1.75\times1.613}{2}+1.27\ =&-\frac{4.096}{3}+\frac{2.823}{2}+1.27\ =&-1.365+1.412 + 1.27\ =&1.317 \end{align*} ]

Answer:

$1.317$