Surface through integral

We are asked to find the area of ​​the passage defined by $y = x^2-8x + 12$ and its tangent lines at points $(0, 12)$ and $(6, 0)$.

I found the points of intersection of the curve with each tangent line ($y_1 = -8x + 12$ and $y_2 = 4x-24$), as well as the point of intersection of the lines and so I came to: $E = E_1 + E_2$, where $E_1$ is the integral of the difference $(y - y_1)$ where $x$ is from $0$ to $3$ and $E_2$ is the integral of the difference $(y - y_2)$ where $x$ is from $3$ to $6$.

However, I came to the conclusion that $E = 0$, having controlled my actions very well. I would appreciate for your guidance.

At the intersection of both tangents,

$y = - 8x+12 = 4x-24 \implies x = 3, y = - 12$

Note that both tangents meet below the parabola. So the integral to find area of the bound region is,

$ =\displaystyle \int_0^3 ((x^2 - 8x + 12) - (-8x+12)) ~ dx ~ + $ $ \displaystyle \int_3^6 ((x^2 - 8x + 12) - (4x-24)) ~ dx$

$ = \displaystyle \int_0^3 x^2 ~ dx + \int_3^6 (6-x)^2 ~ dx = 9 + 9 = 18$