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Why isn't the Riemann integral evaluate in terms of trapeziums ? It would be a better method compared to rectangles.

Isomorphic
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    Why "better"? And could you prove it "better" in general? Besides, rectangles are much easier to deal with. – DonAntonio Dec 31 '13 at 14:05
  • because even though limit will finally give the same answer, but trapeziums with one side flat and one side slant, will give a better approximation and thus it should be more natural to define the integral in terms of trapeziums than rectangles. – Isomorphic Dec 31 '13 at 14:08
  • I can't see how will they give a better approximation than rectangles, but still: it's easier with rectangles, imo. – DonAntonio Dec 31 '13 at 14:09
  • Is it clear (for weird discontinuous functions) that a "trapezium" method converges if and only if the usual "rectangle" method converges? – GEdgar Jan 07 '14 at 15:51

4 Answers4

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They do actually use trapezoids in numerical methods, because they do give better approximation, but the Riemann sum method (with rectangle) is easier to use to prove things. To each his/her own.

Igor Rivin
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The definition of the Riemann integral is a limit. Using trapeziums or rectangles does not change the value of this limit, it is therefore natural to define the integral using the simplest formula based on rectangles.

A difference between rectangles and trapeziums appears when trying to estimate numerically the value of the limit (i.e. the value of the integral), because trapeziums yield a smaller absolute error.

Tom-Tom
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  • Why is the limit taken in terms of rectangles and not trapeziums ? – Isomorphic Jan 07 '14 at 14:22
  • Why doesn't it make a difference ? – Isomorphic Jan 07 '14 at 14:23
  • @Iota: call $I$ the value of the integral (that you want to compute), $R_n$ the integral estimated by means of $n$ rectangles and $T_n$ the integral estaimated by means of $n$ trapeziums. It is known that $|I-R_n|\leq C/n^2$ and $|I-T_n|\leq C/2n^2$, where $C$ depends on the integrated function and the boundaries of the integral. When $n\to\infty$, $R_n$ and $T_n$ both converge to $I$. – Tom-Tom Jan 07 '14 at 14:25
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A Riemann integral is neither evaluated by rectangles nor by trapeziums, but by a certain limiting process involving partitions, Riemann sums, etc. The Fundamental Theorem of Calculus then allows to convert the "infinitesimal" evaluation process into a finite "algebraic" handling of function expressions.

When the exact evaluation of an integral (i.e., expressing its value in terms of known constants) is not possible one has to resort to numerical methods. It is in this realm where it may pay to use trapezoids instead of rectangles in order to obtain more exact approximations with the same amount of numerical work.

Christian Blatter
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This is just a comment to an already overly answered question, which might be useful to other people, who looks at the post.

While this holds for Riemann integrals for the reasons given in the other answers. You cannot assume this for other variants integrals. What I have in mind is stochastic integrals. In this case, the trapezium and the triangle give rises to DIFFERENT notions of integrals/calculus, namely Stratonovich and Ito integrals, respectively.

Moral of the story is that do not assume it is okay in other situations that is not Riemann integration. Certainly not if the integrator is random.

Lost1
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