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I learned that definite integral gives the signed area under a curve by dividing the curve into small rectangular strips and "making" its width shrink to zero.

Using this knowledge summation of certain series can be found. I converted definite integral in this form

$$ \int_{a}^{b} f(x) \, \mathrm{d}x = \lim_{n\to\infty} \frac{b-a}{n} \sum_{r=1}^{n} f\left( a + \left(\frac{b-a}{n}\right)r \right) $$

However, another equivalent form is known to exist, provided below, and I find it much more convenient to use that form. I tried to manipulate the sum to arrive at the equivalent form but so far I've not able to succeed.

$$ \int_{a}^{b} f(x) \, \mathrm{d}x = \lim_{n\to\infty} \frac{1}{n} \sum_{r=g(n)}^{h(n)} f\left(\frac{r}{n}\right), \quad \text{where } \begin{cases} \lim_{n\to\infty} \frac{g(n)}{n} = a, \\ \lim_{n\to\infty} \frac{h(n)}{n} = b. \end{cases} $$

Can someone help (give me a hint perhaps) on how to convert to the second much more convenient expression.

I find second expression much more convenient because the limits of integration can be very easily found using it.

Any help will be appreciated.

EDIT: I've still not able to derive the equivalent relation mentioned above. Can someone hint me?

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  • $\begingroup$ Hmmm.. why isn't someone responding to my question? Sorry if it sounds a bit silly question. $\endgroup$
    – Shivansh J
    Commented Feb 4, 2019 at 9:11
  • $\begingroup$ The notation is a bit unclear to me. For the second sum to make sense, it should hold that $g(n)$ and $h(n)$ are always integers with $g(n) \leq h(n)$ for any $n \in \mathbb N$. $\endgroup$
    – H1ghfiv3
    Commented Mar 5, 2019 at 21:00
  • $\begingroup$ Could you give a practical example for the second expression? $\endgroup$
    – user
    Commented Mar 5, 2019 at 21:07
  • $\begingroup$ @user Consider when lower index of summation is 1 and upper index is 2n. Limits on integration will be given by $\endgroup$
    – Shivansh J
    Commented Mar 5, 2019 at 21:10
  • $\begingroup$ ...a= 1/n (n tends to infinity) = 0 and b = 2n/n (n tends to infinity) = 2 $\endgroup$
    – Shivansh J
    Commented Mar 5, 2019 at 21:11

1 Answer 1

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It more make sense that we focus on the behavior of $$\frac1n\sum_{r=h(n)}^{g(n)}f(r/n)$$ as $n\to \infty$. Think of the $1/n$ as the width of each subinterval of $[a,b]$, and think of $r/n$ as $x$-values in each subinterval. As $n\to\infty$, each subinterval approaches a single point, namely $\lim_{n\to\infty}r/n$. So it's just a sum of $\text{height}\times\text{width}$.

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