# Thread: definite integral....finding the function

1. ## definite integral....finding the function

The following sum
$\frac{1}{1+\frac{3}{n}} \cdot \frac{3}{n} + \frac{1}{1+\frac{6}{n}} \cdot \frac{3}{n} + \frac{1}{1+\frac{9}{n}} \cdot \frac{3}{n} + \ldots + \frac{1}{1+\frac{3 n}{n}} \cdot \frac{3}{n}$
is a right Riemann sum for a certain definite integral $\int_1^b f(x)\, dx$
using a partition of the interval [1,b] into n subintervals of equal length.
i found b to be 4 but i cannot figure out what f(x) is.
i thought it was 1/(1+x) but thats not it =(

2. Originally Posted by mybrohshi5
The following sum
$\frac{1}{1+\frac{3}{n}} \cdot \frac{3}{n} + \frac{1}{1+\frac{6}{n}} \cdot \frac{3}{n} + \frac{1}{1+\frac{9}{n}} \cdot \frac{3}{n} + \ldots + \frac{1}{1+\frac{3 n}{n}} \cdot \frac{3}{n}$
is a right Riemann sum for a certain definite integral $\int_1^b f(x)\, dx$
using a partition of the interval [1,b] into n subintervals of equal length.
i found b to be 4 but i cannot figure out what f(x) is.
i thought it was 1/(1+x) but thats not it =(
I think it is $f(x)=\frac{1}{x}$ in the interval $[1,4]$: they're taking the right extreme points in every subinterval and evaluate there the function: at $1+\frac{3}{n}\,,\,1+\frac{6}{n}\,,...etc.$