Question

Help with Calculus, Riemann Sums.? If someone could help me with the first three I believe i'll be fine from there. http://imgur.com/IBR0Crs

Answer 1

#49

\[\Large \lim_{n \to \infty}\left[\sum_{i=1}^{n}\left(\frac{24i}{n^2}\right)\right]\]

\[\Large \lim_{n \to \infty}\left[\frac{24}{n^2}\sum_{i=1}^{n}\left(i\right)\right]\]

\[\Large \lim_{n \to \infty}\left[\frac{24}{n^2}\left(\frac{n(n+1)}{2}\right)\right] \ \ \ {\text{See Note 1}}\]

\[\Large \lim_{n \to \infty}\left[\frac{12(n+1)}{n}\right]\]

\[\Large \lim_{n \to \infty}\left[\frac{12n+12}{n}\right]\]

\[\Large \lim_{n \to \infty}\left[\frac{12+\frac{12}{n}}{1}\right] \ \ \ {\text{See Note 2}}\]

\[\Large \lim_{n \to \infty}\left[12+\frac{12}{n}\right]\]

\[\Large 12+0 \ \ \ {\text{See Note 3}}\]

\[\Large 12\]

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So,

\[\Large \lim_{n \to \infty}\left[\sum_{i=1}^{n}\left(\frac{24i}{n^2}\right)\right]=12\]


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Note 1: \[\Large \sum_{i=1}^{n}\left(i\right) = \frac{n(n+1)}{2}\] (something to memorize or look up in a table)


Note 2: I divided every term by n (you'll see why in the further steps)


Note 3: I'm using the idean that \[\Large \displaystyle\lim_{n \to \infty} \left(\frac{1}{n^{k}}\right) = 0\] for any positive real number k.