Is my Riemann Sum correct (Example # 2)? 
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:
$$\begin{align*}\Delta x &= \frac7n\\\\
x_i &= -5 + \frac{7i}n\\\\
f(x_i) &= 21 - \frac{70i}{n} + \frac{49i^2}{n^2} \\\\
A&=-738
\end{align*}$$

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