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?
Best Answer
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 $\text{height}\times\text{width}$.