# Thread: Setting up a Riemann sum

1. ## Setting up a Riemann sum

Set up the Riemann sum of f(x)=x^2-2 by partitioning the interval [1,5] into equal subintervals and use the left endpoint of each subinterval for x.

I know the riemann sum in it's general form is
n
Σ f(ci)delta-xi and delta-x = (b-a)/n
i=1

But I have trouble with figuring out the subintervals and using the left endpoint. How do I come up with the left endpoint Ci?

2. The left endpoints are $c_i = a + i\Delta x\;,\;i = 0,1,2 \cdots ,n - 1$.

3. Originally Posted by ebonyscythe
Set up the Riemann sum of f(x)=x^2-2 by partitioning the interval [1,5] into equal subintervals and use the left endpoint of each subinterval for x.

I know the riemann sum in it's general form is
n
Σ f(ci)delta-xi and delta-x = (b-a)/n
i=1

But I have trouble with figuring out the subintervals and using the left endpoint. How do I come up with the left endpoint Ci?
$f(x) = x^2 - 2$
let $P_n := (1=x_0, x_1 , ... , x_{n-1}, x_n =5)$ be a partition into n equal parts..

we have n equal partitions so, using that $\delta - x =\frac{5-1}{n}=\frac{4}{n}$, that is the distance between the partitions, that is, we have
$x_1 = 1 + \frac{4}{n}$
$x_2 = 1 + 2\left( {\frac{4}{n}} \right)$
$x_3 = 1 + 3\left( {\frac{4}{n}} \right)$
up to
$x_{n-1} = 1 + (n-1)\left( {\frac{4}{n}} \right)$
$x_n = 1 + n\left( {\frac{4}{n}} \right) = 5$

so, our $c_i = x_j$ where j = 0, 1, ..., n-1