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Could someone supply a mathematical proof of the following rule:

$$\int_a^b f(x) \ dx \approx (b-a) \left[ \dfrac{f(a)+f(b)}{2} \right]$$

that doesn't rely on partitions as I'm not comfortable with them yet?

Andrew Tawfeek
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Liam
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    http://www.math.ualberta.ca/~etalvila/talks/talvilaboston2.pdf, http://www.math.ucla.edu/~yanovsky/Teaching/Math151A/hw6/Numerical_Integration.pdf – Moo Aug 06 '17 at 23:37
  • There is nothing to add. (+1). – hamam_Abdallah Aug 06 '17 at 23:38
  • Why did this get downvoted? It's a very nice question – étale-cohomology Aug 07 '17 at 00:45
  • @Liam: Could you please clarify what "$\approx$" means here? (The formal answer to your question is "calculate the area enclosed by the secant line through $(a, f(a))$ and $(b, f(b))$", but without a precise definition of the relation symbol it's impossible to be more specific.) – Andrew D. Hwang Aug 07 '17 at 00:51
  • I suspect the word "proof" is causing the downvotes. I would change the word "proof" to "explanation". – littleO Aug 07 '17 at 00:58
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    @étale-cohomology I edited it, but at first it was just a reference to some image on Wikipedia which may have caused the downvotes at the start. – Andrew Tawfeek Aug 07 '17 at 01:05
  • @AndrewD.Hwang They probably intend it to mean approximately, they posted this with a reference to the image containing the formula here: https://en.wikipedia.org/wiki/Trapezoidal_rule – Andrew Tawfeek Aug 07 '17 at 01:06
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    Alright, thank you Andrew! – étale-cohomology Aug 07 '17 at 01:09
  • See this answer https://math.stackexchange.com/a/535304/72031 – Paramanand Singh Aug 07 '17 at 04:36
  • @AndrewTawfeek: Without a definition of "approximately", we have $x \approx y$ for all real $x$ and $y$. (In practice, the integral of $f$ over $[a, b]$ can differ from the right-hand side by an arbitrarily large amount.) – Andrew D. Hwang Aug 07 '17 at 12:03

1 Answers1

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This isn't necessarily a proof per se, but perhaps will shed some insight on why they are approximately equal.

So the area of a trapezoid is given by $\dfrac{b_1+b_2}{2} \cdot h$, where

enter image description here

So now looking at some functions graph and noting the placement of $a,b,f(a),$ and $f(b)$ (where I chose $a$ and $b$ arbitrarily) you can see how the trapezoid begins to form:

enter image description here

and how its area very closely approximates the proper area under the curve (integral). Now using the points that I've labeled on our functions graph and the formula of a trapezoid, the area would come out to be:

$$\dfrac{f(a)+f(b)}{2} \cdot (b-a)$$

(Note that our trapezoid is sort of "on it's side" in the photo, as the height of it is $(b-a)$.)

Now sure, this approximation is great and all, but one obvious problem is the larger our interval is the messier things could get. So one good countermeasure is to break up our entire big interval into smaller intervals, then do the rule above over and over and so on. That way, we get an even better approximation!

So afterwards, things instead look like this:

enter image description here

And of course you can guess that more trapezoids (aka more "partitions") means a better approximation.

Andrew Tawfeek
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