Answer
$\dfrac{9}{2}$
Work Step by Step
We are given that $y=|2x-3| $ $x=0$ to $x=3$
Here, the area $(A)$ can be expressed as: $A=\int_0^{3} |2x-3| \ dx$
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=\int_0^{3} |2x-3| \ dx\\ =[\dfrac{1}{4} (2x-3)|2x-3|]_0^3 $
or, $=[\dfrac{1}{4} (2(3)-3)|2(3)-3|]-[\dfrac{1}{4} (0-3)|0-3|]$
or, $=\dfrac{9}{4}+\dfrac{9}{4}$
Therefore, the required area is: $Area=\dfrac{9}{2}$
