3. What is the area of the region bounded by the curves $y=x^{2003}$ and $y=x^{1 / 2003}$ and lying above the $x$-axis?
Solution:
$1001 / 1002$

The two curves intersect at $(0,0)$ and $(1,1)$, so the desired area is
$$
\int_{0}^{1}\left(x^{1 / 2003}-x^{2003}\right) d x=\left[\frac{x^{2004 / 2003}}{2004 / 2003}-\frac{x^{2004}}{2004}\right]_{0}^{1}=\frac{1001}{1002} .
$$