question let r be the region bounded by the functions f(x)=2x² - 6 and g(x)= - 3x² - 1 as shown in the…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $2x^{2}-6=-3x^{2}-1$. Combine like - terms: $2x^{2}+3x^{2}=-1 + 6$, which gives $5x^{2}=5$, then $x^{2}=1$, and $x=-1,1$.

Step2: Determine the upper - lower functions

On the interval $[-1,1]$, $g(x)\geq f(x)$. The area $A$ between two curves $y = g(x)$ and $y = f(x)$ on $[a,b]$ is given by $A=\int_{a}^{b}[g(x)-f(x)]dx$. Here, $a=-1$, $b = 1$, $g(x)-f(x)=(-3x^{2}-1)-(2x^{2}-6)=-3x^{2}-1 - 2x^{2}+6=-5x^{2}+5$.

Step3: Calculate the definite integral

$A=\int_{-1}^{1}(-5x^{2}+5)dx$. Since $\int(-5x^{2}+5)dx=-\frac{5}{3}x^{3}+5x + C$. Using the fundamental theorem of calculus $\int_{-1}^{1}(-5x^{2}+5)dx=\left(-\frac{5}{3}x^{3}+5x\right)\big|_{-1}^{1}=\left(-\frac{5}{3}(1)^{3}+5(1)\right)-\left(-\frac{5}{3}(-1)^{3}+5(-1)\right)$. $=\left(-\frac{5}{3}+5\right)-\left(\frac{5}{3}-5\right)=-\frac{5}{3}+5-\frac{5}{3}+5=\frac{-5 - 5}{3}+10=\frac{-10}{3}+10=\frac{-10 + 30}{3}=\frac{20}{3}$.

Answer:

$\frac{20}{3}$