Area between the curve $y$, the $x$-axis and the lines 
Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.

I have drawn the graph and concluded that:
$$\int_0^1 0 - (x^2-4x+3)\, dx + \int_1^3 0 - (x^2-4x+3)\, dx = -\frac{-26}{3}$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph? 

$\int_0^1 (x^2-4x+3)\mathbb dx-\int_1^3(x^2-4x+3)\mathbb dx=[(\frac{x^3}3-2x^2+3x)]_0^1-[\frac{x^3}3-2x^2+3x]_1^3=\frac43-[0-\frac43]=\frac83$
