Using integration, what is the area of the triangle bounded by the lines $y = 2x$, $y = \frac{x}{2}$ and $x + y = 6$?

Explanation

1. 01

   Corners: the first two lines meet at the origin; $x + y = 6$ meets $y = 2x$ at $(2, 4)$ and $y = \frac{x}{2}$ at $(4, 2)$.

2. 02

   The upper side changes at $x = 2$: $\int_0^{2} \left(2x - \frac{x}{2}\right)dx + \int_2^{4} \left(6 - x - \frac{x}{2}\right)dx$.

3. 03

   $3 + 3 = 6$ square units.