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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 $f(x) = 2x$?

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  • $\begingroup$ All integrals are of course 100% accurate. But in order to understand what you have computed you have to bring approximations into play. $\endgroup$ – Christian Blatter Jan 1 '16 at 9:04
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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.

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    $\begingroup$ I liked the comparison to the physical world it's just that my initial concern with Riemann sums is that eventually you would need a computer to iterate through all the possible values with the size of dx. while in the geometric world i am used to short precise formulas that give an immediate result $\endgroup$ – soundslikefiziks Jan 1 '16 at 8:50
  • $\begingroup$ Even the classic Pythagorean theorem poses some issues; if we draw two line segments of unit length at precisely $90^\circ$ from each other, then the line segment connecting the two ends should be exactly $\sqrt2$ in length. But can you really measure an irrational length with a physical instrument? My caliper only measures in rational units... $\endgroup$ – Math1000 Jan 1 '16 at 9:00
  • $\begingroup$ it's funny , i couldn't find any b value that would give a rational number in b = sqrt(a^2+a^2). So do you suggest that an anti derivative of x^3/3 + c would provide a sufficient accuracy in calculating the area under the curve instead of going through Riemann sum ? $\endgroup$ – soundslikefiziks Jan 1 '16 at 9:26
  • $\begingroup$ Edit : I meant the anti derivative of x^2 $\endgroup$ – soundslikefiziks Jan 1 '16 at 9:32
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    $\begingroup$ My calipers don't even measure in rational units, or in any exact numbers, but in approximate ranges, limited by the laws of atomic physics. $\endgroup$ – DanielWainfleet Jan 1 '16 at 9:37
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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.

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  • $\begingroup$ So, all integrals are Reimann sums or does integration simply is a symbol of the goal to calculate all values of f(x)dx and Reimann sums are just one application to that goal ? i mean, if i could make my own function to calculate sums would i be still using the same integral symbol and all will be legit ? $\endgroup$ – soundslikefiziks Jan 1 '16 at 5:56
  • $\begingroup$ Riemann sums work for many functions, but not all. 'Lebesgue integrals', and 'measure theory' are used for some more complicated functions. Your own function would have to agree with them. Then you can use the same symbols. $\endgroup$ – Empy2 Jan 1 '16 at 6:07
  • $\begingroup$ So, if you take the trapezoid reimann sum approach and apply it to f(x)=2x you get an 100% accurate answer which is showed in the anti derivative x^2 , but if you apply it on a parabola you only get an approximation , so i was wondering if that's the only case of such accuracy? $\endgroup$ – soundslikefiziks Jan 1 '16 at 6:14
  • $\begingroup$ The integrals that has been proven are 100% accurate, what makes you think they are not? $\endgroup$ – Zelos Malum Jan 1 '16 at 6:16
  • $\begingroup$ @ZelosMalum i did not know about the devision to "proven integrals" and "non proven integrals" , that would certainly make things clearer. everywhere i looked i always saw the wavy equal sign that suggests an approximation of the answer. $\endgroup$ – soundslikefiziks Jan 1 '16 at 6:20

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