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

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 $x$ value of the intersection point.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x = 0$ to the intersection - point $x_0$ is $A=\int_{0}^{x_0}((-2x^{2}+8)-(1.75x + 5))dx=\int_{0}^{x_0}(-2x^{2}-1.75x + 3)dx$.

Step3: Evaluate integral

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(-2x^{2}-1.75x + 3)dx=-\frac{2}{3}x^{3}-\frac{1.75}{2}x^{2}+3x+C$. Evaluate from $0$ to $x_0$: $A=\left(-\frac{2}{3}x_0^{3}-\frac{1.75}{2}x_0^{2}+3x_0\right)-0$. Using a calculator to solve the quadratic equation $2x^{2}+1.75x - 3 = 0$ for the positive root $x_0=\frac{-1.75+\sqrt{(1.75)^{2}-4\times2\times(-3)}}{2\times2}=\frac{-1.75+\sqrt{3.0625 + 24}}{4}=\frac{-1.75+\sqrt{27.0625}}{4}\approx\frac{-1.75 + 5.2}{4}=\frac{3.45}{4}=0.8625$. Then $A=\int_{0}^{0.8625}(-2x^{2}-1.75x + 3)dx=\left[-\frac{2}{3}x^{3}-\frac{1.75}{2}x^{2}+3x\right]_{0}^{0.8625}=-\frac{2}{3}(0.8625)^{3}-\frac{1.75}{2}(0.8625)^{2}+3(0.8625)\approx - \frac{2}{3}(0.6416)-\frac{1.75}{2}(0.7445)+2.5875\approx - 0.4277-0.6524 + 2.5875\approx1.507$.

Answer:

$1.507$