let r be the region bounded by the functions f(x)=4x² - 5 and g(x)=2 as shown in the diagram below. find the…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $4x^{2}-5 = 2$. Then $4x^{2}=7$, $x^{2}=\frac{7}{4}$, $x=\pm\frac{\sqrt{7}}{2}$.

Step2: Set up integral for area

The area $A$ between two curves $y = g(x)$ and $y = f(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}(g(x)-f(x))dx$. Here $a =-\frac{\sqrt{7}}{2}$, $b=\frac{\sqrt{7}}{2}$, $g(x)=2$ and $f(x)=4x^{2}-5$. So $A=\int_{-\frac{\sqrt{7}}{2}}^{\frac{\sqrt{7}}{2}}(2-(4x^{2}-5))dx=\int_{-\frac{\sqrt{7}}{2}}^{\frac{\sqrt{7}}{2}}(7 - 4x^{2})dx$.

Step3: Evaluate the integral

Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(7 - 4x^{2})dx=7x-\frac{4}{3}x^{3}+C$. Then $A=\left[7x-\frac{4}{3}x^{3}\right]_{-\frac{\sqrt{7}}{2}}^{\frac{\sqrt{7}}{2}}=\left(7\times\frac{\sqrt{7}}{2}-\frac{4}{3}\times\left(\frac{\sqrt{7}}{2}\right)^{3}\right)-\left(7\times\left(-\frac{\sqrt{7}}{2}\right)-\frac{4}{3}\times\left(-\frac{\sqrt{7}}{2}\right)^{3}\right)$. Simplify: $A = 2\times\left(7\times\frac{\sqrt{7}}{2}-\frac{4}{3}\times\frac{7\sqrt{7}}{8}\right)=7\sqrt{7}-\frac{7\sqrt{7}}{3}=\frac{14\sqrt{7}}{3}\approx12.124$.

Answer:

$12.124$