Finding the value of the right-endpoint Riemann sum.
2,463
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 $+$.
Related videos on Youtube
Author by
Tony
Updated on January 24, 2020Comments
-
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 almost 4 yearsI got the same as you except for $+$ where you had $-$
-