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

Explanation:

Step1: Find intersection points

Set $-3x^{2}=-6$, then $x^{2} = 2$, so $x=-\sqrt{2}$ and $x = \sqrt{2}$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here $f(x)=-3x^{2}$, $g(x)= - 6$, $a =-\sqrt{2}$, $b=\sqrt{2}$, and $-3x^{2}\geq - 6$ on $[-\sqrt{2},\sqrt{2}]$, so $A=\int_{-\sqrt{2}}^{\sqrt{2}}(-3x^{2}+6)dx$.

Step3: Integrate

$\int(-3x^{2}+6)dx=-x^{3}+6x + C$. Then $\left[-x^{3}+6x\right]_{-\sqrt{2}}^{\sqrt{2}}=(-(\sqrt{2})^{3}+6\sqrt{2})-(-(-\sqrt{2})^{3}+6(-\sqrt{2}))$. $=(-2\sqrt{2}+6\sqrt{2})-(2\sqrt{2}-6\sqrt{2})=8\sqrt{2}\approx11.314$.

Answer:

$11.314$