Answer
$$3$$
Work Step by Step
Here, the area can be expressed as: $A=\int_{0}^{2}[\dfrac{x}{2}-(-x)] \ dx=\dfrac{3}{2} \int_{0}^{2} x \ dx$
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 $\dfrac{3}{2} \int_{0}^{2} x \ dx=\dfrac{3}{2}[\dfrac{x^2}{2}]_{0}^{2}$
or, $=\dfrac{3}{4}[x^2]_0^2$
or, $=\dfrac{3}{4}[4-0]$
Therefore, the required area is: $Area=3$
