Answer
$$2\pi $$
Work Step by Step
$$\eqalign{
& \int_0^2 {x\sqrt {16 - {x^4}} dx} ;{\text{ }} \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}xdx = \frac{1}{2}du \cr
& {\text{The new limits of integration are:}} \cr
& x = 2{\text{ }} \Rightarrow u = {\left( 2 \right)^2} = 4 \cr
& x = 0{\text{ }} \Rightarrow u = {\left( 0 \right)^2} = 0 \cr
& {\text{Apply the substitution}} \cr
& \int_0^2 {x\sqrt {16 - {x^4}} dx} = \int_0^4 {\sqrt {{4^2} - {u^2}} \left( {\frac{1}{2}} \right)} du \cr
& {\text{ }} = \frac{1}{2}\int_0^4 {\sqrt {{4^2} - {u^2}} } du \cr
& {\text{The integral of the form }}\int_0^a {\sqrt {{a^2} - {u^2}} du} ,{\text{ represents a part of a circle}} \cr
& {\text{centered at the origin with radius }}a{\text{ in the first quadrant}},{\text{ then the }} \cr
& {\text{geometry formula}} \cr
& {\text{for the area is }}\int_0^a {\sqrt {{a^2} - {u^2}} du} = \frac{1}{4}\pi {a^2},{\text{ therefore}} \cr
& {\text{ }}\frac{1}{2}\int_0^4 {\sqrt {{4^2} - {u^2}} } du = \frac{1}{2}\left( {\frac{1}{4}\pi {{\left( 4 \right)}^2}} \right) \cr
& {\text{ }} = 2\pi \cr} $$
