## Calculus (3rd Edition)

$\dfrac{128 \sqrt 2}{15} \approx 12.068$
The area under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by: $Area, A= \int_{2}^{4} 2x \sqrt {x-2} \ dx \\ = 2 \int_{2}^{4} x \sqrt {x-2} \ dx \\= 2 \int_2^4 (x-2+2) \sqrt {x-2} \ dx$ Let us apply the substitution method such that: $a=x-2 \implies dx=da$ and $x=a+2$ Now, $A= 2 \int_0^2 (a+2) \times a^{1/2} da \\=\dfrac{4 a^{5/2}}{5}+\dfrac{8 a^{3/2}}{3}]_0^2 \\=\dfrac{128 \sqrt 2}{15} \approx 12.068$