Answer
$\dfrac{39}{4}$ or, $9.75$
Work Step by Step
Here, the area can be expressed as: $A=\int_1^{2}[6x-(x^3-3x)] \ dx=\int_1^{2} (6x-x^3+3x) \ dx$
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 $\int_1^{2} (6x-x^3+3x) \ dx =[3x^2-\dfrac{x^4}{4}+\dfrac{3x^2}{2}]_1^2$
or, $=[3(2)^2-\dfrac{2^4}{4}+\dfrac{3(2)^2}{2}]-[3(1)^2-\dfrac{1}{4}+\dfrac{3(1)^2}{2}]$
or, $=14-\dfrac{17}{4}$
Therefore, the required area is: $Area=\dfrac{39}{4}=9.75$
