Can you interpret any Riemann sum as an infinite number of definite integrals, all giving you the same value when you plug in the limits? For example, this limit of a sum:
$$\lim_{n \to \infty} \frac{3}{n} \left[ 1+ \sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+…+\sqrt{\frac{n}{n+3(n-1)}}\right]$$
Was evaluated in a question as $$\int_0^1(1+3x)^{\frac{-1}{2}}\text{d}x$$
However, by taking out a $\sqrt{3}$ from the denominator of the general expression, i simplified it into $$\sqrt{3}\int_\frac{1}{3}^{\frac{1}{3}+1}\frac{\text{d}x}{\sqrt{x}}$$ and the value is the same, ie., 2.
Is this chance or are multiple possible integrals possible?
Best Answer
The integral that you have written does not equal the limit when you use Reimann Sums, I believe your book is incorrect?
The middle integral is:
$$ \int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x $$
which means you can't factor out a $\sqrt{3}$. So I'll answer the question using the correct Reimann Sum.
Using the Left point method:
$$ A_\mathrm{left} = \Delta x \left[f\left(a\right) + f\left(a + \Delta x\right) + f\left(a+ 2*\Delta x\right) + \cdots + f\left(b - \Delta x\right)\right] $$
$n=\#\ \text{steps}$, $\Delta x = \frac{b-a}{n} = \frac{1}{n}$.
We want to write it in terms of steps rather than step size, so:
$$ \begin{align} A_\mathrm{left} &= \frac{1}{n} \left[f\left(0\right) + f\left(\frac{1}{n}\right) + f\left(2*\frac{1}{n}\right) + \cdots + f\left(1 - \frac{1}{n}\right)\right] \newline &= \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right] \end{align} $$
Therefore:
$$ \int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x = \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right] $$
Which is not really in the nicest form, but I hope this helps!
UPDATE:
Ok so I think I misunderstood the question. So the first limit and last integral are equal to $2$, but the middle integral evaluates to $\frac{2}{3}$ which is not the same. I did the working for the middle integral thinking you were trying to determine the Reimann sum for that, but your initial question had a mistake so rather than deleting everything I will leave the working here in case anyone else finds it helpful, and answer the rest of your question.
Middle integral: $$ \begin{align} \int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x &= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right] \newline &= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \sum_{k=1}^{n-1}\frac{1}{\sqrt{1+3\left(\frac{k}{n}\right)}}\right] \newline &= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \sum_{k=1}^{n-1}\frac{n}{\sqrt{n^2+3kn}}\right] \newline &= \lim_{n \rightarrow \infty} \frac{1}{n} + \lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n-1}\frac{n}{\sqrt{n^2+3kn}} \newline &= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{n+3k}} \newline &=\frac{2}{3} \end{align} $$
First limit: $$ \lim_{n \rightarrow \infty}\frac{3}{n} \sum_{k=0}^{n-1}\sqrt{\frac{n}{n+3k}} = 2 $$
Last integral: $$ \sqrt{3}\int_{\frac{1}{3}}^{\frac{1}{3}+1}\frac{1}{\sqrt{x}}\text{d}{x} = 2 $$
So its clear that the middle and last integral are not the same $2\neq\frac{2}{3}$. But back to your question: Yes certain Reimann Sums may come from multiple integrals.
E.g. If the middle integrals integrand was: $3*f\left(x\right)$, then we would have:
$$ \boxed{3}*\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{n+3k}} $$
Now we use algebra to see if we can get this to the same form as the first limit.
First let the summation index start at $k=0$ (and subtract the $k_0$th term so we don't change the expression):
$$ \begin{align} &=\lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\left[\left(\sum_{\boxed{k=0}}^{n-1}\frac{1}{\sqrt{n+3k}}\right)-\frac{1}{\sqrt{n}}\right] \newline &= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}}-\lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\frac{1}{\sqrt{n}} \newline &= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}} \newline &= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\boxed{\frac{\sqrt{n}}{\sqrt{n}}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}} \newline &= \lim_{n \rightarrow \infty} \frac{3}{n}\sum_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n+3k}} \newline &= \lim_{n \rightarrow \infty} \frac{3}{n}\sum_{k=0}^{n-1}\sqrt{\frac{n}{n+3k}} \newline &= 2 \end{align} $$
As you can see just by modifying the middle integral I was able to reach the first limit!
So I'm going to leave this here and hopefully you can see by this example that you can indeed have multiple integrals linked to a Reiman Sum.