# Is my Riemann Sum correct (Example # 2)? [duplicate]

Possible Duplicate:
Is my Riemann Sum correct?

This is my second attempt, the answer seems rather odd so I thought I would have it checked as well. For the integral: $$\int_{-5}^{2} \left( x^{2} -4 \right) dx$$

My calculations:

$$\begin{align*}\Delta x &= \frac7n\\\\ x_i &= -5 + \frac{7i}n\\\\ f(x_i) &= 21 - \frac{70i}{n} + \frac{49i^2}{n^2} \\\\ A&=-738 \end{align*}$$

-

1
The answer is certainly wrong. How did you get $A$? – Arturo Magidin May 7 '12 at 1:30
But +1 for picking up on the $\LaTeX$. – Brian M. Scott May 7 '12 at 1:47

## marked as duplicate by Peter Tamaroff, Thomas, Noah Snyder, Norbert, draks ...Oct 9 '12 at 18:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

The preliminary computations are fine. That means that the $n$th right hand Riemann sum will be: $$\begin{align*} \text{RHS} &= \sum_{i=1}^n f(x_i)\Delta x\\ &= \sum_{i=1}^n\left(21 - \frac{70i}{n} +\frac{49i^2}{n^2}\right)\frac{7}{n}\\ &= \frac{7(21)}{n}\sum_{i=1}^n1 - \frac{7(70)}{n^2}\sum_{i=1}^n i + \frac{7(49)}{n^3}\sum_{i=1}^ni^2\\ &= \frac{147}{n}(n) - \frac{490}{n^2}\left(\frac{n(n+1)}{2}\right) + \frac{343}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)\\ &= 147 - 245\frac{n^2}{n^2+n} + \frac{343}{6}\frac{n(n+1)(2n+1)}{n^3}, \end{align*}$$ using the formulas that say that $$\begin{align*} 1+2+3+\cdots + n &= \frac{n(n+1)}{2}\\ 1^2+2^2+3^2+\cdots+n^2 &= \frac{n(n+1)(2n+1)}{6}. \end{align*}$$ Now, if we take the limit as $n\to\infty$, we have $$\begin{align*} \lim\limits_{n\to\infty}\frac{n^2}{n^2+n} &= 1\\ \lim\limits_{n\to\infty}\frac{n(n+1)(2n+1)}{n^3} &= 2, \end{align*}$$ which means the area should be $$147 -245 +\frac{343}{3} = -98 + 114+\frac{1}{3} = 16+\frac{1}{3} = \frac{49}{3}.$$

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.952086329460144, "perplexity_flag": "middle"}