the shaded region shown below is bounded by the functions f(x)=-2x² + 8 and g(x)=0.5x + 6, the y - axis and…

Explanation:

Step1: Determine the upper - lower functions

The upper function on the interval $[0,2]$ is $g(x)=0.5x + 6$ and the lower function is $f(x)=-2x^{2}+8$.

Step2: Use the definite - integral formula for the area between two curves

The formula for the area $A$ between two curves $y = g(x)$ and $y = f(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}[g(x)-f(x)]dx$. Here, $a = 0$, $b = 2$, $g(x)=0.5x + 6$ and $f(x)=-2x^{2}+8$. So, $A=\int_{0}^{2}[(0.5x + 6)-(-2x^{2}+8)]dx=\int_{0}^{2}(2x^{2}+0.5x - 2)dx$.

Step3: Integrate term - by - term

We know that $\int(2x^{2}+0.5x - 2)dx=2\times\frac{x^{3}}{3}+0.5\times\frac{x^{2}}{2}-2x+C=\frac{2}{3}x^{3}+\frac{1}{4}x^{2}-2x + C$.

Step4: Evaluate the definite integral

$A=\left[\frac{2}{3}x^{3}+\frac{1}{4}x^{2}-2x\right]_{0}^{2}=\left(\frac{2}{3}\times2^{3}+\frac{1}{4}\times2^{2}-2\times2\right)-\left(\frac{2}{3}\times0^{3}+\frac{1}{4}\times0^{2}-2\times0\right)$. $A=\frac{16}{3}+1 - 4=\frac{16 + 3-12}{3}=\frac{7}{3}\approx2.333$.

Answer:

$2.333$