Answer
$\frac{32}{3}$
Work Step by Step
The integral in question is $\displaystyle\int_0^{16}x^{-1/4}dx$. However, when $x\to0^+$, $x^{-1/4}$ approaches a vertical asymptote. Hence, the integral becomes $\displaystyle\lim_{c\to0^+}\int_c^{16}x^{-1/4}dx=\lim_{c\to0^+}\left[\frac{4}{3}x^{3/4}\right]_c^{16}=\lim_{c\to0^+}\frac{4}{3}(16)^{3/4}-\frac{4}{3}c^{3/4}=\fbox{$\frac{32}{3}$}$ .
