Answer
Improper, converges to 4.
Work Step by Step
The integral has finite bounds so we look at the Definition of Improper Integrals with Infinite Discontinuities.
$f(x)=\displaystyle \frac{1}{\sqrt{x}}$ has an infinite discontinuity at $x=0$, so the case 2 applies:
2. If $f$ is continuous on the interval $(a, b]$ and has an infinite discontinuity at $ a$, then $\displaystyle \int_{a}^{b}f(x)dx=\lim_{c\rightarrow a^{+}}\int_{c}^{b}f(x)dx$.
This is an improper integral.
$\displaystyle \int_{0}^{4}\frac{1}{\sqrt{x}}dx=\lim_{c\rightarrow 0^{+}}\int_{c}^{4}\frac{1}{\sqrt{x}}dx$
$=\displaystyle \lim_{c\rightarrow 0^{+}}\int_{c}^{4}x^{-1/2}dx$
$=\displaystyle \lim_{c\rightarrow 0^{+}}[\frac{1}{1/2}x^{1/2}]_{c}^{4}$
$=\displaystyle \lim_{c\rightarrow 0^{+}}[2\sqrt{x}]_{c}^{4}$
$=2\displaystyle \lim_{c\rightarrow 0^{+}}(2-\sqrt{c})$
$= 2(2-0)$
$=4$
Converges to 4.
