Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 3

$\dfrac{1}{2}$

The area $(A)$ can be calculated as: $A=\int_{0}^1 [(x^3-2x^2+x)-(x^2-x)] \ dx+\int_{1}^2 [(x^2-x)-(x^3-2x^2+x)] \ dx \\= \int_{0}^1 [(x^3-2x^2+x-x^2+x] \ dx+\int_{1}^2 [x^2-x-x^3+2x^2-x] \ dx \\=\int_0^1 [x^3-3x^2+2x] \ dx +\int_1^2 [3x^2-2x-x^3] \ dx \\=[\dfrac{x^4}{4}-x^3+x^2 ]_0^1+[x^3 -x^2-\dfrac{x^4}{4}]_1^2 \\=\dfrac{1}{2}$