Answer
$\dfrac{53}{6}$
Work Step by Step
We are given that $y=|3x-2| $ and $x=0$ to $x=3$
Here, the area $(A)$ can be expressed as: $A=\int_0^{3} |3x-2| \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int |ax+b| \ dx=\dfrac{1}{2a}(ax+b)|ax+b|+C$
Now, we have $A=[\dfrac{1}{(2)(3)} (3x-2)|3x-2|]_0^3 $
or, $=[\dfrac{1}{6} (3x-2)|3x-2|]_0^3$
or, $=[\dfrac{1}{6} (9-2)|9-2|]-[\dfrac{1}{6} (0-2)|0-2|]$
or, $=\dfrac{49}{6}+\dfrac{4}{6}$
Therefore, the required area is: $Area=\dfrac{53}{6}$
