Answer
$$A = \frac{{p - 1}}{{p + 1}},\,\,\,p = 100 \to A = \frac{{99}}{{101}},\,\,\,\,\,p = 1000 \to A = \frac{{999}}{{1001}}$$
Work Step by Step
$$\eqalign{
& \root p \of x \geqslant {x^p}{\text{ on the interval }}\left[ {0,1} \right] \cr
& {\text{The region enclosed is given by}} \cr
& A = \int_0^1 {\left( {\root p \of x - {x^p}} \right)} dx \cr
& A = \int_0^1 {\left( {{x^{1/p}} - {x^p}} \right)} dx \cr
& {\text{Inregrating}} \cr
& A = \left[ {\frac{{{x^{1/p + 1}}}}{{1/p + 1}} - \frac{{{x^{p + 1}}}}{{p + 1}}} \right]_0^1 \cr
& A = \left[ {\frac{{p{x^{\frac{{p + 1}}{p}}}}}{{p + 1}} - \frac{{{x^{p + 1}}}}{{p + 1}}} \right]_0^1 \cr
& A = \left[ {\frac{{p{{\left( 1 \right)}^{\frac{{p + 1}}{p}}}}}{{p + 1}} - \frac{{{{\left( 1 \right)}^{p + 1}}}}{{p + 1}}} \right] - \left[ {\frac{{p{{\left( 0 \right)}^{\frac{{p + 1}}{p}}}}}{{p + 1}} - \frac{{{{\left( 0 \right)}^{p + 1}}}}{{p + 1}}} \right] \cr
& A = \frac{p}{{p + 1}} - \frac{1}{{p + 1}} \cr
& A = \frac{{p - 1}}{{p + 1}} \cr
& \cr
& {\text{Let }}p = 100 \cr
& A = \frac{{100 - 1}}{{100 + 1}} = \frac{{99}}{{101}} \cr
& \cr
& {\text{Let }}p = 1000 \cr
& A = \frac{{1000 - 1}}{{1000 + 1}} = \frac{{999}}{{1001}} \cr} $$
