Are the definite and indefinite integrals actually two different things? Where is the flaw in my understanding? Some context: I'm an engineer, and I tend to have a rather unusual way of understanding and thinking about things, most likely related to my being autistic. I found this question on the HNQ and upon reading it I felt rather confused.

I have always understood that the indefinite and definite integrals are two sides of the same underlying concept, that there is no meaningful difference between them other than that one evaluates to a function and the other to a quantity. Essentially, the definite integral is what you get from running the numbers on the result of an indefinite integral, and saying it's unrelated to the definite integral is like saying that $\sin(x)$ is fundamentally different from $\sin(a)$, assuming $x$ is a variable and $a$ a constant, for no reason other than that $x$ is a variable and $a$ a constant.

To me, whether the definite and indefinite integrals are two sides of the same thing, or two closely related but different things, seems more like a question of philosophy than mathematics.

Where is the flaw in my understanding? Is there one?
The point is, there are three slightly different concepts here:

1) If $f$ and $F$ are functions with the same domain, and $F'(x)=f(x)$ on that domain, then we say that $F$ is an antiderivative of $f$.

2) If $f$ is a function on the interval $a \le x \le b$, then the definite integral of $f$, $\int_a^b f(x) \, dx$, is (loosely speaking) the area under the graph of $y=f(x)$ for $a \le x \le b$, or (more precisely speaking) a limit of Riemann sums. This definition does not involve any derivatives.

3) If $f$ is a function on the interval $a &lt; x &lt; b$ and $c$ is a point in that interval, then the function $G(x)=\int_c^x f(t) \, dt$ could be called an indefinite integral of $f$. This definition of the function $G$ comes directly from the definition in 2) of the definite integral, so it also does not involve any derivatives.

You are saying that you don't see any difference between 2) and 3). You are correct! If your mental definition of "indefinite integral" is 3), then an indefinite integral is just a definite integral with some unknown limits.

The question you're linking is saying that many calculus books use 1) as the definition of an "indefinite integral": that is, they say that if $F$ is an antiderivative of $f$, then $\int f(x) \, dx = F(x)+C$ is the indefinite integral of $f$. They can get away with this because of the fundamental theorem of calculus, which says (in part) that

$$
\frac{d}{dx}\int_c^x f(t) \, dt=f(x) \, .
$$

That is, the function $G(x)$ defined in 3) is an antiderivative of the function $f$, so these two notions of "indefinite integral" are closely related.

But closely related does not mean identical! K B Dave's answer is giving you some examples of situations where the two definitions 1) and 3) diverge slightly. Specifically:


We gave the definition 3) on an interval $a &lt; x &lt; b$, but it doesn't work very well on a more complicated domain. This means that definition 1) is more robust when $f$ has singularities.
The fundamental theorem of calculus says that every "indefinite integral" in the sense of 3) is an antiderivative. It does not say that every antiderivative is an indefinite integral, which in fact is false; you can have functions $f$ and $F$ where $F'(x)=f(x)$, but there is no way to write $F(x)=\int_c^x f(t) \, dt$.
(there are several ways to define integrals, and they are not 100% equivalent, so I'll go the elementary way to try to be clear; for the same reason, I will only discuss continuous functions)

We define the integral of the continuous function $f$ on the interval $[a,b]$ as 
$$\tag1
\int_a^b f(t)\,dt=\lim_{n\to\infty}\sum_{j=1}^n f(x_j)\,\Delta_j,
$$
where $x_j=\frac{j(b-a)}n$ and $\Delta_j=1/n$ (more generally, one uses arbitrary partitions, but it is the same for continuous funcions). 

If you think about it, you will note that there is no trace of a derivative nor  anti-derivative in $(1)$. An integral is an integral, it has a million applications, and if $f$ is nice enough it can be approximated very well by taking $n$ big enough on the right-hand-side of $(1)$ (and maybe tweaking the $x_j$, like using the mid-point rule). 

The relation with derivatives comes from Newton's wonderful Fundamental Theorem of Calculus: if $f$ is a continuous function, then the function 
$$\tag2
F(x)=\int_0^x f(t)\,dt
$$
satisfies $F'(x)=f(x)$. Because of $(2)$, the notation $\int f(t)\,dt$ for an antiderivative of $f$ is so widespread. The fact that any two antiderivatives of $f$ differ by a constant leads to Barrow's Rule: 
$$\tag3
\int_a^b f(t)\,dt=F(b)-F(a). 
$$
So $(3)$ is the reason why we use antiderivatives to calculate integrals (when possible). 

To emphasize the fact that integral means $(1)$, and not $(2)$ nor $(3)$, consider the Error Function
$$
\operatorname{erf}(x)=\frac1{\sqrt\pi}\int_{-x}^x e^{-t^2}\,dt,
$$
which is one of many important functions defined by an integral. The function $\operatorname{erf}(x)$ is a multiple of an  anti-derivative of $f(t)=e^{-t^2}$, and no explicit expression for it  exists.