1. ## Riemann Sum

I was wondering if someone could show me the Riemann Sum for the following function:

$\sqrt{3}x^2 - x + 1$ on [-1,2]

I need to find the exact area under the curve, so I'd assume n->INFINITY

I keep trying this, and I come in the realm of Mathematica's answer (If I split up the interval).. but not close enough.

I keep getting: -1 + 3i/n

Whenever I multiply everything out after substituting into the function, substituting in for i after factoring out the constant/n I come nowhere close.

2. First, we know that the ${\Delta}x = \frac {b-a}{n} = \frac {2-(-1)}{n}= \frac {3}{n}$

Use the right end point rule, we have $C_k = -1 + \frac {3}{n}k$,

Then we would have $C_1= -1 + \frac {3}{n} \rightarrow -1$
and $C_n = -1 + \frac {3}{n}n = 2$

Now, $
f(x) = \sqrt{3}x^2 - x + 1
$
, implies that $f(C_k)=
\sqrt{3}(-1+ \frac {3}{n}k)^2 - (-1+ \frac {3}{n}k) + 1
$

so the Riemann Sum is $\frac {3}{n} \sum ^n _{k=1} \sqrt{3}(-1+ \frac {3}{n}k)^2 - (-1+ \frac {3}{n}k) + 1$

$= \frac {3}{n} \sum ^n_{k=1} \sqrt{3}- \frac {3 \sqrt {3}}{n}k + \frac {9 \sqrt {3}}{n^2}k^2$

3. That's exactly what I've got. Maybe my algebra sucks...

I appreciate your reply. Let me re-work it again and I'll get back here!

$= \sum ^n_{k=1} \sqrt{3}- \frac {3 \sqrt {3}}{n}k + \frac {9 \sqrt {3}}{n^2}k^2$
This is exactly what I have - but, I originally pull the $\frac {3}{n}$ out in front of the summation.

$\frac {3}{n} [ \sum ^n_{k=1} \sqrt{3} - \frac {3 \sqrt {3}}{n} \sum ^n_{k=1} k + \frac {9 \sqrt {3}}{n^2} \sum ^n_{k=1} k^2]$

$\frac {3 \sqrt{3}n}{n} - \frac {9 \sqrt{3}n(n+1)}{2n^2} + \frac {27 \sqrt{3}n(n+1)(2n+1)}{6n^3}$

When I take the limit of that as n->INFINITY, I come nowhere close to

$\frac {3}{2} + 3 \sqrt{3}$

5. I'm not sure where I'm going wrong.. I've tried re-working this quite a few times.

6. Anyone?

7. Let me ask my professor today, I can't figure it out either.