3
$\begingroup$

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:

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

  • 0
    But +1 for picking up on the $\LaTeX$.2012-05-07

1 Answers 1

3

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