A trench is dug beside a wall. A scale drawing uses metres, with the x-axis along ground level…

Step 1. Meeting points. $x^2 - 6x = -2x - 3$ gives $x^2 - 4x + 3 = 0$, so $(x - 1)(x - 3) = 0$.

Step 2. Which is above? At $x = 2$ the plank gives $-7$ and the floor gives $-8$. The plank is above.

Step 3. The area. $\displaystyle\int_1^3 \big((-2x - 3) - (x^2 - 6x)\big)\,dx = \int_1^3 (-x^2 + 4x - 3)\,dx = \left[-\dfrac{x^3}{3} + 2x^2 - 3x\right]_1^3 = 0 - \left(-\dfrac{4}{3}\right) = \dfrac{4}{3}$.