Answer
$\dfrac{1}{3}$
Work Step by Step
Here, the area can be expressed as: $A=\int_{-1}^{0} (-x-x^2) \ dx+\int_{0}^{1} (x-x^2) \ dx=2 \int_{0}^{1} (x-x^2) \ dx$
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 $2 \int_{0}^{1} (x-x^2) \ dx=2[\dfrac{x^2}{2}-\dfrac{x^3}{3}]_{0}^{1}$
or, $=2(\dfrac{1}{2}-\dfrac{1}{3})$
Therefore, the required area is: $Area=\dfrac{1}{3}$
