Answer
$\frac{8}{3}$
Work Step by Step
Step 1. Graph the function for the given interval of $0\leq x\leq 3$ and identify a zero at $x=1$.
Step 2. The total area is the sum of the absolute value of the two separate areas $A_1$ and $A_2$.
Step 3. Evaluate
$A_1=\int_0^1(x^2-4x+3)dx=(\frac{1}{3}x^3-2x^2+3x)|_0^1=\frac{1}{3}(1)^3-2(1)^2+3(1)-0=\frac{4}{3}$
Step 4. Evaluate
$A_2=\int_1^3(x^2-4x+3)dx=(\frac{1}{3}x^3-2x^2+3x)|_1^3=\frac{1}{3}(3)^3-2(3)^2+3(3)-(\frac{1}{3}(1)^3-2(1)^2+3(1))=0-\frac{4}{3}=-\frac{4}{3}$
Step 5. Thus, we have the area as $A=|A_1|+|A_2|=\frac{8}{3}$
