Are all integrals an approximation?

Question

Are all integrals an approximation of the result rather than $100$% accurate ? If so, why is $x^2$ the exact area under the curve of for each value of $f(x) = 2x$?

Answer 1

Much of analysis is fundamentally about "approximations." For example, if a nonnegative real number $x$ satisfies $x<\varepsilon$ for all $\varepsilon>0$, it follows that $x=0$. For if $x$ were greater than zero, we could choose $\varepsilon<x$, yielding a contradiction. So in a sense we can approximate $x$ to arbitrary precision.

To be a bit more concrete, consider the sequence $x_n=\frac1n$, $n=1,2,\ldots$. It is clear that for any $\varepsilon>0$, we may choose a positive integer $N>\frac1\varepsilon$ so that $n\geqslant N$ implies $$|x_n - 0| = \frac1n \leqslant \frac1N<\varepsilon,$$ and so by definition $$\lim_{n\to\infty}x_n = 0. $$ It is not true that $x_n=0$ for any $n$, but we can show that all but finitely many terms of the sequence are as small as we like.

Back to the question of integration - it is possible to physically compute the definite integral of a (integrable) function $f:[a,b]\to\mathbb R$. (Without loss of generality, let us assume $f\geqslant0$) We need only to plot $f$ on a sheet of graph paper with known unit weight, cut the paper along the lines $y=0$, $x=a$, $x=b$, and the curve $\{(x,f(x)):x\in[a,b]\}$ and weight the result. Of course, it is impossible to get an exact result in this method, due to physical constraints (non-uniformity of the paper, finite precision of the scale, only being able to compute and thus plot the function at finitely many points, humidity/air drafts/etc.), but in principle, this is the same idea as a Riemann sum.

Answer 2

Integrals are usually taught as Riemann integrals.
You have upper sums and lower sums. Both these sums are approximations.
But there is usually only one number that is less than all the upper sums, and more than all the lower sums. That makes it the only possible value for the integral - it is exact.