Another Simple Integral

$\begin{aligned} I & = \int_0^3 |3x-5|\ dx \\ & = \int_0^\frac 53 (5-3x)\ dx + \int_\frac 53^3 (3x-5)\ dx \\ & = \left[5x-\frac 32 x^2\right]_0^\frac 53 + \left[\frac 32 x^2-5x\right]_\frac 53^3 \\ & = \frac {25}3 - \frac {25}{6} + \frac {27}2 - 15 - \frac {25}6 + \frac {25}3 \\ & = \frac {100-50+81-90}6 \\ & = \boxed{\dfrac {41}6} \end{aligned}$

Without using Calculus:

The desired area $A = A_{R_{1}} + A_{R_{2}} = \dfrac{1}{2}(\dfrac{5}{3})(5) + \dfrac{1}{2}(\dfrac{4}{3})(4) = \dfrac{25}{6} + \dfrac{16}{6} = \boxed{\dfrac{41}{6}}$ .