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A_1=\int\limits_{0}^{1}(2x^3-6x^2+4x)dx=\left [\frac{1}{2}x^4-2x^3+2x^2\right ] _{0}^{1}

=(\frac{1}{2}-2+2)-0=0,5\ FE

A_2=|\int\limits_{1}^{2}(2x^3-6x^2+4x)dx|=|\left [\frac{1}{2}x^4-2x^3+2x^2\right ] _{1}^{2}|

=|(8-16+8)-0,5|=0,5\ FE

A=A_1+A_2= 1\ FE