question let r be the region bounded by the functions f(x)=3x² and g(x)=5 as shown in the diagram below…

Explanation:

Step1: Find intersection points

Set $3x^{2}=5$, then $x^{2}=\frac{5}{3}$, so $x = \pm\sqrt{\frac{5}{3}}$.

Step2: Set up integral for area

The area $A$ between two curves $y = g(x)$ and $y = f(x)$ is $A=\int_{a}^{b}(g(x)-f(x))dx$. Here $g(x) = 5$, $f(x)=3x^{2}$, $a =-\sqrt{\frac{5}{3}}$, $b=\sqrt{\frac{5}{3}}$. So $A=\int_{-\sqrt{\frac{5}{3}}}^{\sqrt{\frac{5}{3}}}(5 - 3x^{2})dx$.

Step3: Evaluate the integral

Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(5 - 3x^{2})dx=5x-x^{3}+C$. Then $A=\left[5x - x^{3}\right]_{-\sqrt{\frac{5}{3}}}^{\sqrt{\frac{5}{3}}}=\left(5\sqrt{\frac{5}{3}}-\left(\sqrt{\frac{5}{3}}\right)^{3}\right)-\left(-5\sqrt{\frac{5}{3}}+\left(\sqrt{\frac{5}{3}}\right)^{3}\right)=2\left(5\sqrt{\frac{5}{3}}-\frac{5\sqrt{5}}{3\sqrt{3}}\right)=2\times\frac{15\sqrt{5}- 5\sqrt{5}}{3\sqrt{3}}=\frac{20\sqrt{5}}{3\sqrt{3}}=\frac{20\sqrt{15}}{9}\approx8.607$.

Answer:

$8.607$