Finding the value of the right-endpoint Riemann sum.

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Let $x_k=3+\dfrac{9k}n$ and $\Delta x=\dfrac9n$, so

$$\sum_{k=1}^n f(x_k)\Delta x=\sum_{k=1}^n \left(3+\dfrac{9k}n\right)^2\dfrac9n=\sum_{k=1}^n \left(9+\dfrac{54k}n+\dfrac{81k^2}{n^2}\right)\dfrac9n$$

$$=\left(9n+\dfrac{54}n\dfrac{n(n+1)}2+\dfrac{81}{n^2}\dfrac{n(n+1)(2n+1)}6\right)\dfrac9n,\\$$

which approaches $(9+27+27)9=567$ as $n$ approaches $\infty$.


Your answer also approaches $567$ as $n\to\infty,$ but you had $-$ where I have $+$.

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Tony
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Updated on January 24, 2020

Comments

  • Tony
    Tony almost 4 years

    Calculate the area between $𝑓(𝑥)=𝑥^2$ and the x axis over the interval [3,12] using a limit of right-endpoint Riemann sums:
    Find the value of the right-endpoint Riemann sum in terms of n:
    $$\sum_{k=1}^n f(x_k)\Delta x=$$
    I got $81+\frac{243(n-1)}{n}+\frac {729(n-1)(2n-1)}{(6n^2)}$ but it comes up as wrong.

    • J. W. Tanner
      J. W. Tanner almost 4 years
      I got the same as you except for $+$ where you had $-$