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anonymous

  • one year ago

Need help with Riemann sum equation

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  1. anonymous
    • one year ago
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    \[\int\limits_{1}^{6}4x-2dx\] n=3, so using 3 rectangles and left endpoints

  2. SolomonZelman
    • one year ago
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    If you want the rectangles to be even, \(\Delta x=\dfrac{b-a}{n}=\dfrac{6-1}{3}=\dfrac{5}{3}\) So your intervals \([x_0,~~~x_1]\) \([x_1,~~~x_2]\) \([x_2,~~~x_3]\) (And \([x_1=x_0+\Delta x]\) ) are going to be: \([1,~~~1+5/3]\) ---> \([1,~~~8/3]\) \([8/3,~~~8/3+5/3]\) ---> \([8/3,~~~13/3]\) \([13/3,~~~13/3+5/3]\) ---> \([13/3,~~~6]\)

  3. SolomonZelman
    • one year ago
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    You are doing *left* endpoints, and therefore your first rectangle will have the height of *f(1)*. Then, the second rectangle will have a height of *f(8/3)* and the third rectangle will have a height of *f(13/3)*.

  4. SolomonZelman
    • one year ago
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    So basically, your area by reinman sums: assuming that \(\Delta x\) is same (which I found to be 5/3) for all intervals (as I supposed in the beginning), *AND* with the appropriate heights (which I listed accordingly) \(\large\color{black}{ \displaystyle {\rm A}=\frac{5}{3}f(1)+\frac{5}{3}f\left(\frac{8}{3}\right)+\frac{5}{3}f\left(\frac{13}{3}\right) }\) Adding three rectangles. \(\Uparrow\) And then after factoring you get: \(\large\color{black}{ \displaystyle {\rm A}=\frac{5}{3}\color{red}{\left\{\color{black}{f(1)+f\left(\frac{8}{3}\right)+f\left(\frac{13}{3}\right)}\right\}} }\)

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