Area enclosed by lines and axes

The region is bounded by: $x = 0$ (Y-axis), $x = 2$, $y = 0$ (X-axis), $y = |x - 2| + 4$.

For $x \in [0, 2]$, $x - 2 \le 0$, so $|x - 2| = 2 - x$. Hence $y = (2 - x) + 4 = 6 - x$.

The shape is a trapezium with vertices $(0,0)$, $(2,0)$, $(2,4)$, and $(0,6)$. Area = $\int_{0}^{2} (6 - x) dx = \left[ 6x - \frac{x^2}{2} \right]_0^2 = 12 - 2 = 10$.