the shaded region shown below is bounded by the functions f(x)=-2x² + 8 and g(x)=1.75x + 5 and the x and y…

Explanation:

Step1: Find intersection point

Set $-2x^{2}+8 = 1.75x + 5$. Rearrange to $2x^{2}+1.75x - 3=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b=1.75$, $c=-3$, we get the positive root $x = 1$.

Step2: Set up integral for area

The area $A=\int_{0}^{1}((1.75x + 5)-0)dx+\int_{1}^{2}((-2x^{2}+8)-0)dx$. For $\int_{0}^{1}(1.75x + 5)dx=\left[\frac{1.75}{2}x^{2}+5x\right]{0}^{1}=\frac{1.75}{2}+5=\frac{1.75 + 10}{2}=5.875$. For $\int{1}^{2}(-2x^{2}+8)dx=\left[- \frac{2}{3}x^{3}+8x\right]_{1}^{2}=\left(-\frac{16}{3}+16\right)-\left(-\frac{2}{3}+8\right)=\left(\frac{-16 + 48}{3}\right)-\left(\frac{-2 + 24}{3}\right)=\frac{32}{3}-\frac{22}{3}=\frac{10}{3}\approx3.333$.

Step3: Calculate total area

$A = 5.875+\frac{10}{3}\approx5.875 + 3.333=9.208$.

Answer:

$9.208$