Say you have the curve $y=x^3-5x^2+6x$ with roots 0, 2 and 3. Integrating $y$ I got the equation $\frac14x^4-\frac53x^3+3x^2+C$. I found the value of $C$ by subbing a point form the curve such as $(0,2)$ and found the value for $C$ to be $-\frac83$, after which I found that $\int_0^2y\,dx=\frac83$ and $-\int_2^3y\,dy=\frac5{12}$. Adding them up gave the area as $\frac{37}{12}$, however the answer is $\frac72$.
I am not entirely sure if the method and my answer is correct, or if the answer given is incorrect?
you will have the integral $$\int_{0}^{2}(x^3-5x^2+6x)dx+\left|\int_{2}^{3}(x^3-5x^2+6x)dx\right|$$ can you solve this?