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I am a calculus 1 student. I was wondering that if Riemann sums only give us an approximation(either over-estimate or under-estimate) the area under the curve, Why do we celebrate Riemann sums(considering it came out in the 19th century) when we can actually find the exact area using integrals.

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    $\begingroup$ Beause you can't always compute the exact integral. $\endgroup$ – Thomas Andrews Apr 11 at 20:53
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    $\begingroup$ Most integrals are hopeless. Numerical methods are all we have. $\endgroup$ – lulu Apr 11 at 20:53
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    $\begingroup$ Also, Riemann sums are the simplest way to define "integrals" exactly, via successive Riemann sums. We call the (most basic) integral the Riemann integral for a reason. (There's another definition, using Lebesque's definition, but it is more involved.) $\endgroup$ – Thomas Andrews Apr 11 at 20:54
  • $\begingroup$ In later courses, you will need them to prove some theorems using the sums. Also, what happens when an integral is impossible to do? Furthermore, the series make the idea concrete. $\endgroup$ – Bor Kari Apr 11 at 20:57
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    $\begingroup$ When you do definite integrals, you're using the fundamental theorem of calculus and that requires antiderivatives, but not all functions have antiderivatives. The most famous example of a function that has no elementary antiderivative is $$\int e^{-x^2}\,dx.$$ $\endgroup$ – Michael Rybkin Apr 11 at 21:00
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1) Definitions.

A Riemann Integral is by definition a precise limit of Riemann sums. One can use this to justify which functions are (or aren't) integrable, and to further derive the two fundamental theorems of calculus. All subsequent properties of Riemann Integrals (substitution, breaking into a sum of two integrals, etc.) are in fact derived through looking at limits of Riemann Sums.

2) Not all functions have an explicit anti-derivative.

There are tons of functions that do not have an explicit anti-derivative. These either occur in some compact form like $\exp(-x^2)$, or more generally tend to arise as data measurements from some experiment. In the latter case, the measurements are discrete approximations of a non-discrete function.

So the second-best approach is to approximate the integral via Riemann Sums. More precisely, Quadratures are used, as vanilla Riemann Sums might not converge quickly.

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The Riemann sums are used to construct the integral, to define the object. When the functions to be integrated are "nice enough" you have learned a simple formula to compute the integral (involving primitives), but this rule does not define the integral, nor does it allow to compute every integral.

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The Riemann sums definition of the integral are used in the fundamental theorem of calculus to prove that the anti-derivative is the area under the curve.

If all you care about is "getting the answer" then you can forget about most of what you learned in the Riemann sum discussion.

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