$a^x-a$ is non-positive for x in [0,1]. $a^x$ is continuous at 1 so given **any** $\epsilon>0$, there is a d in (0,1) such that $|a^x-a|<\epsilon$ for all x in [d,1]. $\int_0^1 a^x x^a =\int_0^d (a^x-a) x^a+\int_d^1 (a^x-a) x^a+ \int_0^1 ax^a$
The first integral say $I_1\ge \int_0^d d^a (a^x-a)\ge\int_0^d d^a -\int_0^d ad^a\to 0$ as $a\to \infty$.