Answer
$2$
Work Step by Step
Here, the area can be expressed as: $A=\int_{0}^{1} [(x^2-2x) -(-x^2+4x-4)] \ dx+\int_{1}^{2} [(-x^2+4x-4) -(x^2+-2x)] \ dx$
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 $\int_{0}^{1} (2x^2-6x+4) \ dx+\int_{1}^{2} (-2x^2+6x-4) \ dx=[\dfrac{2x^3}{3}-3x^2+4x]_{0}^{1}+[-\dfrac{2x^3}{3}+3x^2-4x]_1^2$
or, $=[\dfrac{2(1)^3}{3}-3(1)^2+4(1)]-0+[-\dfrac{2(2)^3}{3}+3(2)^2-4(2)]-[-\dfrac{2(1)^3}{3}+3(1)^2-4(1)]$
or, $=\dfrac{5}{3}-0-\dfrac{4}{3}+\dfrac{5}{3}$
Therefore, the required area is: $Area=2$
