# Is area under an integral limit exact or an approximation?

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.

• It is exact. If not then the symbol $=$ is not used. – drhab Nov 3 '14 at 15:42

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 $0$.

• What about $lim_{x \to 2}\frac{x^2-4}{x-2}=4$? We do say here the the limit approaches 4 as x approaches 2. – user117913 Nov 3 '14 at 16:26
• no, we say that the function approaches $4$ as $x$ approaches $2$. That is at least correct, but still confusing. I would say "the limit of the function is $4$ at $x=2$". That helps to dispel the false believe that limits are somehow changing, approximating, whateverating. – Ittay Weiss Nov 3 '14 at 16:44

The integral gives you 'exactly' the right area in the sense that any approximation, using rectangles or squares--which have a well defined area--will always converge on the answer you find evaluating the integral.

In other words, the way we define what the area under the curve in introductory analysis of a Riemann sum is tautologically equal to the integral.

Do you have another definition for 'the exact area' under the curve? (There's another definition called a Lebesgue integral and it turns out that for continuous functions like sin it is exactly equal to the Riemann integral.)

• But for a curve, as in Riemann sums, the curve is broken into infinitesimally small pieces. However still there must be still some area left uncovered by the infinitesimally thin rectangle. So how does the integral remove this error? – user117913 Nov 3 '14 at 15:49
• Also, you didn't answer the question related to limits. – user117913 Nov 3 '14 at 15:49
• You can show that for a continuous function like sin that as the approximation gets better and better, as say the number of partitions of the interval $[x_1, x_2]$ goes to infinity, that the error becomes arbitrarily small. – Simon S Nov 3 '14 at 15:50
• There is an error, then? – user117913 Nov 3 '14 at 15:52
• @user117913 There are no infinitesimal real numbers, so an "infinitely small" uncovered area is zero. There is no such thing in the real numbers as an infinitely small leftover. The integral is the limit, where the error is zero. Note that you can use other shapes than rectangles (trapezoids, for example) that better approximate the curve at intermediate steps, and you get the same answer in both cases. If there were some "infinitely small" error that came from using rectangles the answers should differ by a significant fraction of that error (and at any finite width, they do). – Robert Mastragostino Nov 3 '14 at 15:54

This is exactly why we have the concept of limits.

We have a curve, $f(x) = \sin(x)$ from $0 \to \pi$

$\displaystyle \int_{0}^{\pi} \sin(x) \space dx = \displaystyle \lim_{||\Delta(x)|| \to 0}\sum_{i=1}^{n} f(c_i)\Delta(x)$ where $c_i$ is a respective point in the respective partition $i$.

It is saying, as the width of the Largest partition approaches $0$, what will the area be? Thinking about this, if the width of the largest partition $=0$ then that is the same as $n \to \infty$, now, if the number of rectangles goes to infinity, then the rectangles have width $dx$ making the rectangle infinitely small in width. This means the whole region is covered perfectly.

Again this is a limit, this cannot physically happen. Which is why we put it as a limit, and don't just say as $n \to \infty$, it is the limit. What would happen if $n \to \infty$

Does this help?

• Could you further explain why limits so nearly wipe out all errors? And does the accuracy have a bound? – user117913 Nov 3 '14 at 15:54
• @user117913, it wipes out all errors, because it is essentially saying that if $n \to \infty$ then that would happen. So it is saying let $n = \infty$. There is no room for error. – Amad27 Nov 3 '14 at 15:58
• One should never write $n=\infty$ in this context. It is nonsensical since $\infty$ is not a natural number. – Ittay Weiss Nov 3 '14 at 16:21