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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:

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

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marked as duplicate by Pedro Tamaroff, Thomas, Noah Snyder, no identity, draks ... Oct 9 '12 at 18:21

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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

1 Answer 1

up vote 3 down vote accepted

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 $$147 -245 +\frac{343}{3} = -98 + 114+\frac{1}{3} = 16+\frac{1}{3} = \frac{49}{3}.$$

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