let r be the region bounded by the functions f(x)=-3x² + 2 and g(x)=-10 as shown in the diagram below. find…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $-3x^{2}+2 = - 10$. Rearranging gives $3x^{2}=12$, then $x^{2}=4$, and $x=\pm2$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ where $f(x)\geq g(x)$ on $[a,b]$ is $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $a=-2$, $b = 2$, $f(x)=-3x^{2}+2$ and $g(x)=-10$. So $A=\int_{-2}^{2}[(-3x^{2}+2)-(-10)]dx=\int_{-2}^{2}(-3x^{2}+12)dx$.

Step3: Use integral rules

Since $\int(-3x^{2}+12)dx=-x^{3}+12x + C$, and $\int_{-2}^{2}(-3x^{2}+12)dx=\left[-x^{3}+12x\right]_{-2}^{2}$.

Step4: Evaluate definite - integral

$(-2^{3}+12\times2)-(-(-2)^{3}+12\times(-2))=(-8 + 24)-(8-24)=16 - (-16)=32$.

Answer:

$32$