let r be the region bounded by the functions f(x)=-2x² + 6 and g(x)=4x² as shown in the diagram below. find…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $- 2x^{2}+6 = 4x^{2}$. Combine like - terms: $6x^{2}=6$, then $x^{2}=1$, and $x=-1,1$.

Step2: Set up the integral for the 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, $f(x)\geq g(x)$ on $[-1,1]$, so $A=\int_{-1}^{1}((-2x^{2}+6)-4x^{2})dx=\int_{-1}^{1}(6 - 6x^{2})dx$.

Step3: Integrate

Use the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$. Then $\int(6 - 6x^{2})dx=6x-2x^{3}+C$.

Step4: Evaluate the definite integral

$A=\left[6x-2x^{3}\right]_{-1}^{1}=(6\times1-2\times1^{3})-(6\times(-1)-2\times(-1)^{3})=(6 - 2)-(-6 + 2)=4-(-4)=8$.

Answer:

$8$