Why is $\int_0^3 \frac{1}{\sqrt{x-3}}\, \mathrm{d}x$ a complex number? This is my first question on this site, so please pardon any nuanced formatting errors. 

My friend and I were discussing the following integral yesterday: 

$$\int_0^3 \frac{1}{\sqrt{x-3}} \, \mathrm{d}x$$

I discovered that the integral evaluates to $-2i\sqrt{3}$ This confuses me. When I studied calculus 1, I was taught that the definite integral can be thought of as a way to calculate the "area under the curve" or the accumulation of a value. How can an area have a complex value? 

Thus, I know that my conceptual understanding of integrals is flawed. How can I reconcile the concept of "area" with a complex value? Or, what is a "better" (for lack of a better word) way to conceptualize the integral? 

Thanks in advance!
The area under the curve idea applies to real-valued functions. This function is imaginary for almost all values of the area of integration.
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  Your problem is 'well defined' once you specifies the $\ds{\root{z - 3}}$ branch-cut. For instance,
  $$
\root{z - 3} =
\root{\verts{z - 3}}\exp\pars{{\,\mathrm{arg}\pars{z - 3} \over 2}\,\ic}
\,,\quad
z \not= 3\,,\quad
0 &lt; \,\mathrm{arg}\pars{z - 3} &lt; 2\pi
$$
  and the branch cut is along  the 'line segment'
  $$
\braces{z = 3 + r\expo{\ic\alpha}\,,\quad r \geq 0\,,\quad
0 \leq \alpha &lt;\pi\quad \mbox{or}\quad \pi &lt; \alpha &lt; 2\pi}
$$ 



In that case,
\begin{align}
&amp;\color{#f00}{%
\int_{0}^{3}{\dd x \over \root{\verts{x - 3}}\expo{\ic\pars{\pi - \alpha}}/2}} =
-\ic\expo{\ic\alpha/2}\int_{0}^{3}{\dd x \over \root{3 - x}} =
\color{#f00}{-2\root{3}\expo{\ic\alpha/2}\,\,\ic}
\end{align}


  The result $\ds{-2\root{3}\,\ic}$ is given for a branch-cut along the 'positive real axis'. Namely,$\ds{\braces{z = x + 0\ic\,,\quad x \geq 3}}$ which is equivalent to $\ds{\alpha = 0}$.