Answer
$$1$$
Work Step by Step
Here, the area can be expressed as: $A=\int_{-1}^{0} (x^2-x) \ dx+\int_{0}^{1} (x-x^2) \ dx$
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 $\int_{-1}^{0} (x^2-x) \ dx+\int_{0}^{1} (x-x^2) \ dx=[\dfrac{x^3}{3}-\dfrac{x^2}{2}]_{-1}^{0}+[\dfrac{x^2}{2}-\dfrac{x^3}{3}]_0^1$
or, $=-[-\dfrac{1}{3}-\dfrac{1}{2}]+[\dfrac{1}{2}-\dfrac{1}{3}]$
or, $=\dfrac{5}{6}+\dfrac{1}{6}$
Therefore, the required area is: $Area=1$
