# Weird sum problem

• November 20th 2009, 03:32 PM
Velvet Love
Weird sum problem
Compute the numerical value of the Riemann sum Sum(F(x[k])(x[k]-x[k-1]),k = 1,k=3) for the integral: (integral range) [3,7] ∫ of F(x)*ⅆx where F(x)=x+3. Assume that the points x[0] .. x[3] divide the interval [3,7] into 3 equal subintervals.
• November 21st 2009, 04:47 AM
Scott H
Here is a hint: the subintervals are

$\left[3,4\frac{1}{3}\right]\;\;\;\;\;\left[4\frac{1}{3},5\frac{2}{3}\right]\;\;\;\;\;\left[5\frac{2}{3},7\right]$

and the value of $x_k-x_{k-1}$ in each case is $1\frac{1}{3}$. The Riemann sum is therefore

\begin{aligned}
\sum_{k=1}^3 f(x_k)(x_k-x_{k-1})&=f\left(4\frac{1}{3}\right)\cdot 1\frac{1}{3}+f\left(5\frac{2}{3}\right)\cdot 1\frac{1}{3}+f(7)\cdot 1\frac{1}{3}\\
&=1\frac{1}{3}\cdot\left(f\left(4\frac{1}{3}\right )+f\left(5\frac{2}{3}\right)+f(7)\right).
\end{aligned}