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Math Help - Weird sum problem

  1. #1
    Junior Member
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    Sep 2009
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    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.
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    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}<br />
\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}\\<br />
&=1\frac{1}{3}\cdot\left(f\left(4\frac{1}{3}\right  )+f\left(5\frac{2}{3}\right)+f(7)\right).<br />
\end{aligned}
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