Answer
$\dfrac{20}{3}$
Work Step by Step
Here, the area can be expressed as: $A=\int_0^{2}[0-(x^3-4x^2)] \ dx=\int_0^{2} (4x^2-x^3) \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Now, we have $\int_0^{2} (4x^2-x^3) \ dx=[\dfrac{4x^3}{3}-\dfrac{x^4}{4}]_0^2$
or, $=[\dfrac{4(2)^3}{3}-\dfrac{(2)^4}{4}]-[\dfrac{4(0)^3}{3}-\dfrac{(0)^4}{4}]$
or, $=\dfrac{32}{3}-4$
Therefore, the required area is: $Area=\dfrac{20}{3}$
