Answer
$\dfrac{32}{3}$
Work Step by Step
We are given that $y=4-x^2 $ and $x=-2$ to $x=2$
Here, the area $(A)$ can be expressed as: $A=\int_{-2}^{2} (4-x^2) \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Now, we have $A=[4x-\dfrac{x^3}{3}]_{-2}^2 $
or, $=[4(2)-\dfrac{2^3}{3}]-[4(-2)-\dfrac{(-2)^3}{3}]$
or, $=\dfrac{16}{3}+\dfrac{16}{3}$
Therefore, the required area is: $Area=\dfrac{32}{3}$
