let r be the region bounded by the functions f(x)=3x² - 7 and g(x)= - 3x² + 8 as shown in the diagram below…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $3x^{2}-7=-3x^{2}+8$. Combine like - terms: $3x^{2}+3x^{2}=8 + 7$, $6x^{2}=15$, $x^{2}=\frac{15}{6}=\frac{5}{2}$, $x=\pm\sqrt{\frac{5}{2}}$.

Step2: Determine the upper and lower functions

On the interval $[-\sqrt{\frac{5}{2}},\sqrt{\frac{5}{2}}]$, $g(x)\geq f(x)$. The area $A$ between two curves $y = g(x)$ and $y = f(x)$ is given by $A=\int_{a}^{b}[g(x)-f(x)]dx$, where $a =-\sqrt{\frac{5}{2}}$, $b=\sqrt{\frac{5}{2}}$, $g(x)-f(x)=(-3x^{2}+8)-(3x^{2}-7)=-6x^{2}+15$.

Step3: Calculate the integral

$A=\int_{-\sqrt{\frac{5}{2}}}^{\sqrt{\frac{5}{2}}}(-6x^{2}+15)dx$. Since the integrand $y=-6x^{2}+15$ is an even function, we can rewrite it as $A = 2\int_{0}^{\sqrt{\frac{5}{2}}}(-6x^{2}+15)dx$. Integrating term - by - term: $\int(-6x^{2}+15)dx=-6\times\frac{x^{3}}{3}+15x=-2x^{3}+15x$. Then $A = 2\left[-2x^{3}+15x\right]_{0}^{\sqrt{\frac{5}{2}}}=2\left(-2\left(\sqrt{\frac{5}{2}}\right)^{3}+15\sqrt{\frac{5}{2}}\right)$. Using a calculator: [ \begin{align*} A&=2\left(-2\times\frac{5\sqrt{10}}{4}+15\times\frac{\sqrt{10}}{2}\right)\ &=2\left(-\frac{5\sqrt{10}}{2}+\frac{15\sqrt{10}}{2}\right)\ &=2\times\frac{10\sqrt{10}}{2}\ &=10\sqrt{10}\approx 31.623 \end{align*} ]

Answer:

$31.623$