Limit of a Riemann Sum as an Integral

Question

The teacher assigned us Limits of Riemann sums to transform into their integral form only (we have not learned the FTC and thus cannot evaluate them using anti derivatives so we are just rewriting)

This one in particular was brought up and confused everyone:

$\lim _{n\to \infty }\left(\frac{1}{n}\left(\frac{1}{1}+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\frac{1}{1+\frac{\left(n-1\right)}{n}}\right)\right)$

Some of us got:

\begin{equation} \int _0^1\left(\frac{1}{1+x}\right)dx\: \end{equation}

Others got: \begin{equation} \int _1^2\left(\frac{1}{x}\right)dx\: \end{equation}

When ran through my Ti-84 calculator, both integrals come out to be the exact same numerical answer. Who is correct? Or is there more than one answer to this?

Answer 1

Both are correct since $\displaystyle\int_{0}^{1}\dfrac{1}{1+x}dx=\ln(1+x)\bigg|_{x=0}^{x=1}=\ln 2$ and $\displaystyle\int_{1}^{2}\dfrac{1}{x}dx=\ln x\bigg|_{x=1}^{x=2}=\ln 2$.

Answer 2

In the first integral substitute $y=1+x$.

Answer 3

$\frac{1}{1+x}$ is just $\frac{1}{x}$ shifted over $1$ space to the left, so this means in order for the integrals to be the same we need to add one to the bounds of $\frac{1}{x+1}$ to get the integral for $\frac{1}{x}$.