Consider the function $\displaystyle f(x) = -\frac{x^2}{2} + 6}$

I need to calculate $\displaystyle \displaystyle \int_0^3 ({-\frac{x^2}{2} + 6})dx$ by using the definition of Riemann sum for $\displaystyle R_n $ - right end-points,

$\displaystyle \sum_{n=1}^{n} (-\frac{x^2}{2} + 6)$

so, I started by finding $\displaystyle \delta x = \frac{3-0}{n} = \frac{3}{n}$

after that

$\displaystyle a) R_n = \sum_{n=1}^{n} (-\frac{(\frac{3i}{n})^2}{2} + 6) (\frac{3}{n}) = $

$\displaystyle b) R_n = \sum_{n=1}^{n} (-\frac{(\frac{9i^2}{n^2})}{2} + 6) (\frac{3}{n}) = $

$\displaystyle c) R_n = \sum_{n=1}^{n} (-\frac{9i^2}{2n^2} + 6) (\frac{3}{n}) = $

$\displaystyle d) R_n = \sum_{n=1}^{n} (-\frac{9i^2 + 12n^2}{2n^2}) (\frac{3}{n}) = $

$\displaystyle e) R_n = \sum_{n=1}^{n} (-\frac{3(3i^2 + 4n^2)}{2n^2}) (\frac{3}{n}) = $

$\displaystyle f) R_n =(\frac{3}{n}) (-\frac{3}{2n^2})\sum_{n=1}^{n} [(3i^2 + 4n^2)] = $

$\displaystyle g) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{3(n(n+1)(2n+1)}{6} + 4n^2)] = $

$\displaystyle h) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1)}{2} + 4n^2)] = $

$\displaystyle i) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1) + 8n^2}{2}] = $

so, I'm basically asked to provide

$\displaystyle R_n = $ as a function of n, without summation symbols, and for this I entered step i) , but it was considered as wrong, so obviously, my mistake is somewhere above,

$\displaystyle \lim_{n\to infty} R_n$, here I obtained $\displaystyle -\frac{9}{2}$ , but it is actually wrong since all my work-up before is wrong,

can you, please help me with this,

thanks,

dokrbb