TITLE: Is area under an integral limit exact or an approximation?
QUESTION [7 upvotes]: Suppose we need to calculate the area of the the curve $y=sin x$. Then we calculate the area  enclosed by the curve from $x=x_1$ to $x=x_2$ as $\int_{x_1}^{x_2}sin x\, dx$. Is the area calculated so exact or approximate?. In case of a linear curve such $y=ax+b$, we do get an exact value (as justified by geometry)
This confusion stems from the question whether limits are exact or not (and hence all operations related to limits).
Note: I did see other posts to see an answer, but none of them were complete or the questions were something else entirely. So this is not a duplicate.

REPLY [5 votes]: The notion of area must first be defined. Unfortunately, that is quite a tricky business. In measure theory one gives a precise notion of area for sets in $\mathbb R^2$ but it turns out that not every subset can meaningfully be assigned an area. However, it is a theorem that if a set is precisely the locus of points bounded above and below by graphs of functions $f(x)$ and $g(x)$ respectively, then the area of that set is $\int f(x)-g(x)dx$. This is then a precise justification for computing the area under the graph of a function by means of the integral. It is then a precise answer. 
Remark: You seem to be under the impression that limits are somehow imprecise or that they are approximations. This is incorrect. A limit is a number. It is not a process, nor an approximation, nor in any way imprecise. It is a very much fixed number that never ever changes. The limit $\lim _{n\to \infty }\frac{1}{n}$ is precisely $0$. It is wrong to say "it is $0$ when $n=\infty $" since $n$ is a natural number here, so $n=\infty $ is meaningless. It is wrong to say "the limit approaches $0$" or "the limit becomes $0$" or any other thing like that. The limit is simply $0$.