text: # Weird sum problem
  
  • Nov 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.
  • Nov 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}