Part two would be done as so
I = \int_{0}^{2}x^3\sqrt{4-x^2}dx
= \int_{0}^{2}x^2 \times x \sqrt{4-x^2}dx
u^2 = 4 - x^2 \Rightarrow 2udu = -2xdx \Rightarrow udu = -xdx
(x = 0, u = 2), (x = 2, u = 0)
\text{Now rearranging the original Integral we can see how this u sub fits in}
I =...