Just Find Its Antiderivative Right?

I know that most of you chose the answer of 2, because that was what your teacher told you. Specifically, that

$\int_0^1 \frac{1}{ \sqrt{x} } \, dx = [ 2 \sqrt{x} ]_0^1 = 2$

However, you are not performing the Riemann integral when you evaluate this way! The Riemann-Lebesgue theorem states that

The function $f: [ a , b ] \rightarrow \mathbb{R}$ is Riemann integrable if and only if (it is bounded and the set of points at which it is discontinuous has Lebesgue measure zero).

I know that most of you learnt that the riemann sums is calculated as a left sum and as a right sum, and if the sums converge then the Riemann integral exists. However, they neglected to explain the criteria under which we only need to consider left and right sums. There are various criteria on the functions, like having a bounded+continuous function, or having a monotonic function on a compact interval.

As an explicit example, we know that the function

$f(x) = \begin{cases} 0 & x \in \mathbb{Q} \\ 1 & x \not \in \mathbb{Q} \\ \end{cases}$ is not Riemann-integrable. However, if we simply took the intervals with endpoints $\frac{ k}{n}$ , we would calculate that the left and right Riemann sums are always 0. We see that this function is neither bounded+continuous, nor monotonic, and hence we cannot just consider left and right sums.

Similarly, in the case of $\frac{1}{ \sqrt{x} }$ , because the function isn't defined at 0, the function isn't continuous at 0. On the other hand, if we tried to fix this by removing 0 from the domain, we do not have a compact interval (even though the function would be monotonic). As such, we once again cannot just consider left and right sums.

Now, let's see why the bounded criteria is necessary in the theorem, by working from first principles of the Riemann integral. The Riemann integral of $f$ equals $S$ if the following condition holds:

For all $\epsilon > 0$ , there exists a $\delta$ such that for any tagged partition $x_0, \ldots , x_n$ and $t_0, \ldots , t_{n-1}$ whose mesh is less than $\delta$ , we have
$| \sum_{i=0}^n{n-1} f(t_i) ( x_{i+1} - x_i) - S | < \epsilon$

If the function is unbounded and tends to (positive) infinity at the point $p = 0$ , then given any partition, we can always find a sequence of tags such that $x_i \leq p \leq x_{i+1}$ and $t_{ i_n}$ with $t_{i_n} \rightarrow + \infty$ . With this sequence of tags, we see that $\sum_{i=0}^{i=n-1} f(t_i) ( x_{i+1} - x_i)$ is unbounded, and thus the Riemannian integrability condition cannot hold. Thus, all unbounded functions are not Riemann-integrable.

The entire Riemann-Lebesgue theorem takes a bit more machinery to prove. Give it a shot!

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Note: For those people who try and justify the initial equation using the fundamental theorem of calculus , once again we have to note the conditions under which the theorem holds. In particular, we need that $f$ is continuous on the closed interval $[a,b]$ , which isn't true for $\frac{1}{\sqrt{x} }$ as it is not defined at $x = 0$ , and removing that point leaves us an open interval (which is not compact).