question let r be the region bounded by the functions f(x)=-4x² and g(x)=-4 as shown in the diagram below…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $-4x^{2}=-4$. Solving for $x$ gives $x^{2} = 1$, then $x=-1$ and $x = 1$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $a=-1$, $b = 1$, $f(x)=-4x^{2}$ and $g(x)=-4$. So $A=\int_{-1}^{1}(-4x^{2}-(-4))dx=\int_{-1}^{1}(4 - 4x^{2})dx$.

Step3: Integrate

Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(4 - 4x^{2})dx=4x-\frac{4x^{3}}{3}+C$.

Step4: Evaluate definite integral

$A=\left[4x-\frac{4x^{3}}{3}\right]_{-1}^{1}=\left(4\times1-\frac{4\times1^{3}}{3}\right)-\left(4\times(-1)-\frac{4\times(-1)^{3}}{3}\right)=(4-\frac{4}{3})-(-4+\frac{4}{3})=( \frac{12 - 4}{3})-(\frac{-12 + 4}{3})=\frac{8}{3}-\left(-\frac{8}{3}\right)=\frac{16}{3}$.

Answer:

$\frac{16}{3}$