# Thread: help with Riemann sum (RHS) of Definite Inegral

1. ## help with Riemann sum (RHS) of Definite Inegral

Could somone show me how the integral X^2-7X+10 on [2,5] would be equal to -4.5. I know that given RHS= F(Xi)delta x the delta is 3/n and Xi is 1+3i/n so I know how to start like this.

(1+(3i/n)^2-7(1+(3i/n))+10)(3/n)

I just don't see how I would break this down.

I know that after it is simplifed I need to take the limit as n approaches postive infintiy.

2. The algebra is the booger in these things more than anything.

${\Delta}x=\frac{3}{n}$

Using the right end method.

$x_{k}=a+k{\Delta}x$

$x_{k}=2+\frac{3k}{n}$

$\left[\left(2+\frac{3k}{n}\right)^{2}-7\left(2+\frac{3k}{n}\right)+10\right]\cdot\frac{3}{n}$

Now, expand out:

$\frac{27k^{2}}{n^{3}}-\frac{27k}{n^{2}}$

Use the formulae for the sum of the squares and integers.

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

Now, take the limit:

$\lim_{n\to \infty}\frac{9}{2n^{2}}-\lim_{n\to \infty}\frac{9}{2}$

Now, you can see the 4.5...right?.