let r be the region bounded by the functions f(x)=-3x² and g(x)=3x² - 6 as shown in the diagram below. find…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $- 3x^{2}=3x^{2}-6$. Rearranging gives $6x^{2}=6$, then $x^{2} = 1$, and $x=-1,1$.

Step2: Determine upper - lower functions

On the interval $[-1,1]$, $g(x)\geq f(x)$. 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 $a=-1$, $b = 1$, $g(x)=3x^{2}-6$ and $f(x)=-3x^{2}$. So $A=\int_{-1}^{1}[(3x^{2}-6)-(-3x^{2})]dx=\int_{-1}^{1}(6x^{2}-6)dx$.

Step3: Use integral rules

Since $\int(6x^{2}-6)dx=6\times\frac{x^{3}}{3}-6x = 2x^{3}-6x+C$. By the fundamental theorem of calculus, $A=\left[2x^{3}-6x\right]_{-1}^{1}$.

Step4: Evaluate definite integral

$A=(2\times1^{3}-6\times1)-(2\times(-1)^{3}-6\times(-1))$. $A=(2 - 6)-(-2 + 6)$. $A=-4-4=-8$. But area is non - negative, so $A = 8$.

Answer:

$8$