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

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $-2x^{2}-6x=x^{2}+3x$. Rearrange to $3x^{2}+9x = 0$, factor out $3x$: $3x(x + 3)=0$. The solutions are $x=-3$ and $x = 0$.

Step2: Determine upper - lower function

On the interval $[-3,0]$, $f(x)\geq g(x)$. The area $A$ between two curves $y = f(x)$ and $y = g(x)$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $a=-3$, $b = 0$, and $f(x)-g(x)=(-2x^{2}-6x)-(x^{2}+3x)=-3x^{2}-9x$.

Step3: Calculate the integral

$A=\int_{-3}^{0}(-3x^{2}-9x)dx=\left[-x^{3}-\frac{9}{2}x^{2}\right]_{-3}^0$. First, substitute $x = 0$: $-(0)^{3}-\frac{9}{2}(0)^{2}=0$. Then substitute $x=-3$: $-(-3)^{3}-\frac{9}{2}(-3)^{2}=27-\frac{81}{2}=\frac{54 - 81}{2}=-\frac{27}{2}$. Take the absolute value, $A=\frac{27}{2}$.

Answer:

$\frac{27}{2}$