question let r be the region bounded by the functions f(x)=4x² and g(x)= - 4x² + 7 as shown in the diagram…

Explanation:

Step1: Find intersection points

Set $4x^{2}=-4x^{2}+7$. Then $8x^{2}=7$, so $x^{2}=\frac{7}{8}$, and $x = \pm\frac{\sqrt{7}}{\sqrt{8}}=\pm\frac{\sqrt{14}}{4}$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ is $A=\int_{a}^{b}[g(x)-f(x)]dx$, where $a =-\frac{\sqrt{14}}{4}$, $b=\frac{\sqrt{14}}{4}$, $g(x)=-4x^{2}+7$ and $f(x)=4x^{2}$. So $A=\int_{-\frac{\sqrt{14}}{4}}^{\frac{\sqrt{14}}{4}}((-4x^{2}+7)-4x^{2})dx=\int_{-\frac{\sqrt{14}}{4}}^{\frac{\sqrt{14}}{4}}(7 - 8x^{2})dx$.

Step3: Evaluate integral

Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$ and $\int kdx=kx + C$ ($k$ is a constant), we have $\int(7 - 8x^{2})dx=7x-\frac{8x^{3}}{3}+C$. Then $A=\left[7x-\frac{8x^{3}}{3}\right]_{-\frac{\sqrt{14}}{4}}^{\frac{\sqrt{14}}{4}}=\left(7\times\frac{\sqrt{14}}{4}-\frac{8}{3}\times\left(\frac{\sqrt{14}}{4}\right)^{3}\right)-\left(7\times\left(-\frac{\sqrt{14}}{4}\right)-\frac{8}{3}\times\left(-\frac{\sqrt{14}}{4}\right)^{3}\right)=2\times\left(7\times\frac{\sqrt{14}}{4}-\frac{8}{3}\times\frac{14\sqrt{14}}{64}\right)=2\times\left(\frac{7\sqrt{14}}{4}-\frac{7\sqrt{14}}{12}\right)=2\times\frac{21\sqrt{14}-7\sqrt{14}}{12}=2\times\frac{14\sqrt{14}}{12}=\frac{7\sqrt{14}}{3}\approx8.082$.

Answer:

$8.082$