Answer
$\dfrac{16}{3}$
Work Step by Step
The area is given by $A=\int_0^{2}[0-(x^2-4)] \ dx=\int_0^{2} (4-x^2) \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Here, we have $\int_0^{2} (4-x^2) \ dx=[4x-\dfrac{x^3}{3}]_0^2$
or, $=[4(2)-\dfrac{2^3}{3}]-[4(0)-\dfrac{(0)^3}{3}]$
or, $=8-\dfrac{8}{3}$
Therefore, the required area is: $Area=\dfrac{16}{3}$
