Help with integrals and Riemann sums.

Question

Find $f$, such that

$$\sum_{i=1}^n \frac{1}{n(2+\frac{i}{n})\ln(2+\frac{i}{n})}$$

is a Riemann sum of $f$ on the interval $[0,1]$.

Also, how is this connected to the integral of $f$?

Thanks for the help.

Answer 1

You can get a "Riemann sum" approximating the integral $\int_a^b f(x) dx$ by dividing the distance from a to b into n intervals (so each interval has length $\Delta x= (b- a)/n$ and evaluating $f(x_i)$ where $x_i$ is some value of x inside the $i^{th}$ interval. If we take it at the left end of the interval, it is $a+ \Delta x$. Multiply each $f(x_i)$ by $\Delta x$ and sum: $\sum_{i= 0}^n f(x_i)\Delta x$.

Seeing that "n" in the denominator I would take $\Delta x= \frac{1}{n}$ and $x_i= 2+ \frac{i}{n}$. As i goes from 0 to n, $x_i$ goes from $2+ \frac{0}{n}= 2$ to $2+ \frac{n}{n}= 3$ so the integral corresponding to this Riemann sum is $\int_2^3 \frac{1}{xln(x)}dx$.

That can easily be integrated using the substitution u= ln(x).