Thomas' Calculus 13th Edition

$1000$
Let $f(r)=\int_{0}^{1} \dfrac{1}{r^{0.999}} dr$ Now, $\lim\limits_{k \to 0^{+}} f(r)= \lim\limits_{k \to 0^{+}}\int_{0}^{1} \dfrac{1}{r^{0.999}} dr$ Thus, $\lim\limits_{k \to 0^{+}} [\dfrac{r^{0.001}}{0.001}]_{1}^{(k)}=\lim\limits_{k \to 0^{+}} [1000-1000(k)^{-0.001}]=1000-0= 1000$