# definite integral....finding the function

• Nov 25th 2009, 01:50 PM
mybrohshi5
definite integral....finding the function
The following sum
$\displaystyle \frac{1}{1+\frac{3}{n}} \cdot \frac{3}{n} + \frac{1}{1+\frac{6}{n}} \cdot \frac{3}{n} + \frac{1}{1+\frac{9}{n}} \cdot \frac{3}{n} + \ldots + \frac{1}{1+\frac{3 n}{n}} \cdot \frac{3}{n}$
is a right Riemann sum for a certain definite integral $\displaystyle \int_1^b f(x)\, dx$
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 =(
• Nov 25th 2009, 01:58 PM
tonio
Quote:

Originally Posted by mybrohshi5
The following sum
$\displaystyle \frac{1}{1+\frac{3}{n}} \cdot \frac{3}{n} + \frac{1}{1+\frac{6}{n}} \cdot \frac{3}{n} + \frac{1}{1+\frac{9}{n}} \cdot \frac{3}{n} + \ldots + \frac{1}{1+\frac{3 n}{n}} \cdot \frac{3}{n}$
is a right Riemann sum for a certain definite integral $\displaystyle \int_1^b f(x)\, dx$
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 $\displaystyle f(x)=\frac{1}{x}$ in the interval $\displaystyle [1,4]$: they're taking the right extreme points in every subinterval and evaluate there the function: at $\displaystyle 1+\frac{3}{n}\,,\,1+\frac{6}{n}\,,...etc.$