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I teach A Level Maths in the UK. We are required to do some 'introduction' to integral from first principles as part of the specification (link, page 25 is the interesting part).

In a previous exam question (Paper 2, June 2018), this was essentially the question:

Suppose we have the curve $y= \sqrt{x}$

The point $P(x,y)$ lies on the curve. Consider a rectangle with height $y$ and width $\delta x$.

Calculate $\displaystyle \lim_{\delta x \rightarrow 0}\sum_{x = 4}^9 \sqrt{x} \,\delta x$

The answer involves us recognising $$ \lim_{\delta x \rightarrow 0}\sum_{x = 4}^9 \sqrt{x} \, \delta x = \int_4^9 \sqrt{x} \, dx$$ and evaluating the integral.

Is this notation standard?

To me, it doesn't make sense. How can you have $x=4$ to $9$ as the limits on the sum, for example? A sum only works over integral values.

In addition, one could easily give meaning to the limit as $\displaystyle \lim_{\delta x \rightarrow 0}\sum_{x = 4}^9 \sqrt{x} \, \delta x = \lim_{\delta x \rightarrow 0}( \sqrt{4}(\delta x) + \sqrt{5} (\delta x) + \cdots + \sqrt{9} (\delta x))$ which should be $0$ as $\delta x \rightarrow 0$.

The reason I am so confused is that this question and notation appears on a Pearson A Level Maths Exam paper. It is a regulated qualification. There are very qualified people out there that have deemed this to make sense and be used to assess the understanding of thousands of A Level students in the UK.

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  • $\begingroup$ The question is from an A Level exam paper, which is the public exams students aged 16-18 in the UK sit before going to university. As a result I am confused since one would think that it has some credibility. $\endgroup$ – PhysicsMathsLove Oct 18 '20 at 15:43
  • $\begingroup$ I wish so, but unfortunately the mark scheme used against these students requires the association between the quoted sum and the integral. $\endgroup$ – PhysicsMathsLove Oct 18 '20 at 15:45
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    $\begingroup$ This is a question testing content on page 25 of the A Level Maths specification under Integration: qualifications.pearson.com/content/dam/pdf/A%20Level/… $\endgroup$ – PhysicsMathsLove Oct 18 '20 at 15:50
  • $\begingroup$ No news or errors to find - it is approved by the Department for Education, hence I come onto stackexchnage to be enlightened for how this is valid. $\endgroup$ – PhysicsMathsLove Oct 18 '20 at 15:51
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    $\begingroup$ Ryan G mentioned in a comment that you mentioned in a question three years ago that you've seen this notation in many places in books and online, which potentially settles your own question. Even if a UK mathematician like J.G. wouldn't use this notation, it's not just limited to Pearson, by your own discovery. $\endgroup$ – Mark S. Oct 19 '20 at 13:05
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  1. $$\lim_{\delta x \to0}\sum_{r = 1}^n f(x_r^*)\,\delta x$$ and $$\lim_{n \to\infty}\sum_{r = 1}^n f(x_r^*)\,\delta x$$ are equivalent expressions (containing Riemann sums).

  2. Loosely generalising the index of summation to $x$, $$\lim_{\delta x \to0}\sum_{x = x_1^*}^{x_n^*} f(x)\,\delta x,$$ I reckon that the expression remains conceptually sound.

  3. If the limits of integration are $a$ and $b$, $$\lim_{\delta x \to0}\sum_{x = a}^{b-\delta x} f(x)\,\delta x$$ (containing a right Riemann sum) is equivalent to the above expression.

  4. $$\lim_{\delta x \to0}\sum_{x = a}^{b} f(x)\,\delta x$$ converges to the same value, I think. This is the notation used in the syllabus/specification in question.

    However, the sum now ceases to be a Riemann sum, as we are now considering $(n+1)$ instead of just $n$ subintervals. I don't find this hand-waving beneficial to thoughtful learners, because it conflicts with the intuitive partition of $n$ rectangles of (in our case fixed) width $\delta x$.

Given that the specification (p. 25) requires only that the candidate recognise the limit of a sum as an integral, and the exam (Q5) accordingly just expects the candidate to rewrite $\displaystyle \lim_{\delta x \to 0}\sum_{x = 4}^9 \sqrt{x} \, \delta x\,\,\text{ as }\,\displaystyle \int_4^9\sqrt{x} \,\mathrm{d}x$ $\,\,\big(\text{for 1 mark}\big),$ I think the nonstandard notation is acceptable (albeit not particularly instructive).

ADDENDUM

$\delta x$ is the width of the $n$ subintervals of our regular partition, i.e., $\displaystyle\delta x=\frac{b-a}n$, so $\delta x$ and $n$ vary in tandem as the Riemann sum approaches the definite integral.

Your error in $\boxed{\lim_{\delta x \rightarrow 0}\sum_{x = 4}^9 \sqrt{x} \, \delta x = \lim_{\delta x \rightarrow 0}\left( \sqrt{4}(\delta x) + \sqrt{5} (\delta x) + \cdots + \sqrt{9} (\delta x)\right)=0}$ was in choosing a step size of 1 instead of letting it vary as $\delta x$. (After all, with the index of summation generalised from $r$ to $x$, the step size ought to be correspondingly changed from $1$ to $\delta x$.) This is why the first equality in the box is false.

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    $\begingroup$ "I reckon that the expression remains conceptually sound." This answer would be improved with a non-Pearson source using them summation symbol with non-integer indices in that sort of way (especially in 3 or 4). I'm in the US and none of 2-4 would be acceptable (though the error in 2 is more clear) and am curious if this is a Pearson problem or a genuine UK difference. $\endgroup$ – Mark S. Oct 19 '20 at 11:11
  • $\begingroup$ @MarkS. To be clear, it would have been patently false to claim that $\displaystyle\sum_{x = a}^{b}\big(f(x)\,\delta x\big)$ equals $\displaystyle\sum_{r=1}^n\big(f(x_r^*)\,\delta x\big)$ or even $\displaystyle\sum_{x=x_1^*}^{x_n^*}\big(f(x)\,\delta x\big)$. Pearson, while not being rigorous, made no such claim in the two citations. $\endgroup$ – Ryan G Oct 19 '20 at 11:43
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    $\begingroup$ I feel that by asking students to calculate the limit of the first and expecting them to evaluate the limit of the last, they implicitly did make such a claim. (Unless your position is that "calculate" has a meaning on the exam that isn't a refinement of "write an equal expression"?) $\endgroup$ – Mark S. Oct 19 '20 at 11:54
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    $\begingroup$ @MarkS. Helpfully, in the OP's related question from 3 yrs ago, he observed that the offending expression " is given in my school books and a lot of online websites I see as the definition of the definite integral." Apparently, this notation isn't isolated to Pearson. $\endgroup$ – Ryan G Oct 19 '20 at 12:05
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    $\begingroup$ @MarkS.This specification (p. 42) from OCR (a competing exam board also in the UK) uses the exact same offending notation. $\endgroup$ – Ryan G Oct 19 '20 at 14:57
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In my experience as a UK mathematician, this isn't standard notation. The aim is to convey $x$'s step size is $\delta x$ rather than $1$. I'd recommend writing the limit as $\lim_{\delta x\to0}\sum_{k=0}^{5/\delta x}\sqrt{4+k\delta x}\delta x$ - no, wait, make it $\lim_{N\to\infty}\sum_{k=0}^N\sqrt{4+5k/N}(5/N)$. This limit is the stated integral; in fact, if we set the maximum $k$ to $N-1$ instead (which won't change the limit), it's a Riemann sum.

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    $\begingroup$ "To my mind, this isn't standard notation." Forgive me if your identity/location is well known elsewhere on the site, but it would add value to your answer if you provided your relevant authority. I feel the same way as in your post, but I'm only a moderately experienced US Calculus teacher, rather than, say, someone who has taught in the UK for 50 years and helped design similar exams for companies other than Pearson, etc. $\endgroup$ – Mark S. Oct 19 '20 at 11:14
  • $\begingroup$ @MarkS. Does my edit help? $\endgroup$ – J.G. Oct 19 '20 at 11:52
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Just change the equals sign to an arrow. Altogether, it should sufficiently demonstrate the intention to change a discrete sum to a more accurate continuous sum. The equals sign is indeed improper, so a simple arrow addresses this without exceeding A-Level needs. It is also reasonable ink usage, for a generally trivial incongruity.

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  • $\begingroup$ Why the downvote? This is a good observation. $\endgroup$ – K.defaoite Oct 19 '20 at 11:31
  • $\begingroup$ @K.defaoite I did not downvote, but this appears to miss the point of the question. If "the equals sign is indeed improper", then we can't really solve the problem with Pearson's exams just by "[changing] the equals sign to an arrow". Rather, the OP has to tell their students "if you ever see this on a Pearson exam, misinterpret it in the following way (and write an arrow just in case) or else you may lose marks". To confirm that "the equals sign is indeed improper", it would help if some authority (other exams in the UK, personal authority as a UK instructor?) were added to this answer. $\endgroup$ – Mark S. Oct 19 '20 at 11:51
  • $\begingroup$ With regard to the use of an arrow (which from my experience is common practice at A-Level), it is the sum on the left which is inaccurate, which an arrow doesn't really fix. I have seen many A-Level teachers write similar sum to integral equations, but the better instructors may have pointed out the inaccuracy to their students. I know a Chief Examiner of one of the UK boards, so I will ask him what he thinks. It might just be best to omit the discrete sum and write a few words, followed by an arrow to the integral. $\endgroup$ – Dr Mathelogique Oct 22 '20 at 0:11

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