I was trying to convert a Riemann Sum to an integral however I got the parameters wrong according to the answers from the textbook. Now the Riemann Sum involves a $\ln$ as the function as is as follows:
$$\sum_{i=1}^n {2\over n} \ln ( 1 + {2i\over n})$$
I assumed that the $\operatorname{dx}$ would be equal to $\frac{2}{n}$ so then $\frac{b-a}{n} = \frac{2}{n}$. Then I assumed that $x_i$ would be $1 + 2\frac{i}{n}$ so where $a$ is equal to $1$ and $b$ is equal to $3$ to produce $2$ from the $\frac{2}{n}$. However the answers in my textbooks say that the function is that of $\ln(1+x)$ where the parameters are $b = 2$ and $a = 0$. How do I know which one to use?
Ok, what you have is probably $$ S=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{2}{n} \log \bigg(1+\frac{2k}{n} \bigg) $$ here clearly $a=1,b=2+1=3$. Hence you get the integral $$ S=\int_{1}^{3}\log x dx $$ EDIT: the integrand can also be $\log(1+x)$, and then the bounds are $a=0, b=2$