Answer
$\frac{1}{2}$
Work Step by Step
Step 1. Given the function $y=x^3-3x^2+2x=x(x-1)(x-2)$ and the interval $[0,2]$, find the zeros at $x=0,1,2$; thus we need to divide the integration into two regions: $[0,1],[1,2]$
Step 2. We have $A_1=\int_{0}^{1}(x^3-3x^2+2x)dx=(\frac{1}{4}x^4-x^3+x^2)|_{0}^{1}=(\frac{1}{4}(1)^4-(1)^3+(1)^2)-(\frac{1}{4}(0)^4-(0)^3+(0)^2)=\frac{1}{4}$
Step 3. Similarly, $A_2=\int_{1}^{2}(x^3-3x^2+2x)dx=(\frac{1}{4}x^4-x^3+x^2)|_{1}^{2}=(\frac{1}{4}(2)^4-(2)^3+(2)^2)-(\frac{1}{4}(1)^4-(1)^3+(1)^2)=-\frac{1}{4}$
Step 4. The total area is given by $A=|A_1|+|A_2|=\frac{1}{2}$
