question find the area of the shaded region below. do not round, enter an exact answer. f(x)=x^{2}-3 g(x)=1…

Explanation:

Step1: Identify the upper - and lower - functions

The upper function is $g(x)=1$ and the lower function is $f(x)=x^{2}-3$.

Step2: Find the intersection points

Set $1=x^{2}-3$, then $x^{2}=4$, so $x = - 2$ and $x = 2$. These are the limits of integration.

Step3: Use the area formula

The area $A$ between two curves $y = g(x)$ and $y = f(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[g(x)-f(x)]dx$. Here, $a=-2$, $b = 2$, $g(x)=1$ and $f(x)=x^{2}-3$. So $A=\int_{-2}^{2}[1-(x^{2}-3)]dx=\int_{-2}^{2}(4 - x^{2})dx$.

Step4: Integrate

We know that $\int(4 - x^{2})dx=4x-\frac{1}{3}x^{3}+C$. Then $\int_{-2}^{2}(4 - x^{2})dx=\left[4x-\frac{1}{3}x^{3}\right]_{-2}^{2}$.

Step5: Evaluate the definite integral

$\left(4\times2-\frac{1}{3}\times2^{3}\right)-\left(4\times(-2)-\frac{1}{3}\times(-2)^{3}\right)=(8-\frac{8}{3})-(-8+\frac{8}{3})=16-\frac{16}{3}=\frac{32}{3}$.

Answer:

$\frac{32}{3}$