Answer
$\displaystyle \frac{3\cdot 5^{2/3}}{2}$
(converges)
Work Step by Step
The integrand has a discontinuity at 5, so this is a type 2.a improper integral.
$\displaystyle \int_{a}^{b}f(x)dx=\lim_{t\rightarrow b^{-}}\int_{a}^{t}f(x)dx$
$I=\displaystyle \int_{0}^{5}\frac{1}{\sqrt[3]{5-x}}dx=\lim_{t\rightarrow 5^{-}}\int_{0}^{t}\frac{1}{\sqrt[3]{5-x}}dx$
$\left[\begin{array}{ll}
u=5-x & \\
du=-dx & \\
x=0\Rightarrow u=5, & x=5\Rightarrow u=0
\end{array}\right]$
$I=\displaystyle \lim_{u\rightarrow 0^{-}}\int_{-5}^{u}-u^{-1/3}dx=\lim_{u\rightarrow 0^{-}}\left[-\frac{3u^{2/3}}{2}\right]_{-5}^{u}$
$=0-(-\displaystyle \frac{3\cdot 5^{2/3}}{2})$
$=\displaystyle \frac{3\cdot 5^{2/3}}{2}$
(converges)
