# definite integral answer different from Riemann Sum calc

• May 18th 2008, 07:15 PM
rexexdesign
definite integral answer different from Riemann Sum calc
Hi everybody,

I have to calculate a definite integral using the oh so necessary Riemann Sum.
I first calculated the integral with the fundamental theorem as a check.
The integral is from 1 to 2 of (2+3x^2)dx and it equals 9 according to my calculation.

When I calculated with tv Riemann sum, it equaled 1, so I must have made a mistake.

I started by finding delta x = 1/n and xi = i/n
then I set up the equation to be:
lim of n->inf Sigma i=1 to n of f(i/n) 1/n

Isn't that correct?
I then solved it and got 1 as the answer.

If anybody could help me out that would be great.

Thx
• May 18th 2008, 07:37 PM
Mathstud28
Quote:

Originally Posted by rexexdesign
Hi everybody,

I have to calculate a definite integral using the oh so necessary Riemann Sum.
I first calculated the integral with the fundamental theorem as a check.
The integral is from 1 to 2 of (2+3x^2)dx and it equals 9 according to my calculation.

When I calculated with tv Riemann sum, it equaled 1, so I must have made a mistake.

I started by finding delta x = 1/n and xi = i/n
then I set up the equation to be:
lim of n->inf Sigma i=1 to n of f(i/n) 1/n

Isn't that correct?
I then solved it and got 1 as the answer.

If anybody could help me out that would be great.

Thx

$\lim_{n\to\infty}\sum_{x=1}^{n}f(M_i)\Delta{x_i}$

so $\Delta{x_i}=\frac{2-1}{n}$

$f(M_i)=2+3\bigg(1+\frac{x}{n}\bigg)^2=5+\frac{6x}{ n}+\frac{3x^2}{n^2}$

So $\frac{1}{n}\sum_{x=1}^{\infty}\bigg[5+\frac{6x}{n}+\frac{6x^2}{n^2}\bigg]=\frac{1}{n}\bigg[5n+\frac{6n(n+1)}{2n}+\frac{3n(n+1)(2n+1)}{6n^2}\b igg]=$ $\frac{1}{n}\bigg[5n+3n+\frac{3}{n}+n+\frac{3}{2}+\frac{1}{2n}\bigg]=5+3+\frac{3}{n^2}+1+\frac{3}{2n}+\frac{1}{2n^2}$
The limit of that as it goes to infinity is 9

$\lim_{n\to\infty}\sum_{x=1}^{n}\bigg[\bigg(2+3\bigg(1+\frac{x}{n}\bigg)^2\bigg)\bigg(\f rac{1}{n}\bigg)\bigg]=9=\int_{1}^{2}\bigg[2+3x^2\bigg]dx$

(Whew)(Whew)(Whew)
• May 18th 2008, 07:45 PM
TheEmptySet
Quote:

Originally Posted by rexexdesign
Hi everybody,

I have to calculate a definite integral using the oh so necessary Riemann Sum.
I first calculated the integral with the fundamental theorem as a check.
The integral is from 1 to 2 of (2+3x^2)dx and it equals 9 according to my calculation.

When I calculated with tv Riemann sum, it equaled 1, so I must have made a mistake.

I started by finding delta x = 1/n and xi = i/n
then I set up the equation to be:
lim of n->inf Sigma i=1 to n of f(i/n) 1/n

Isn't that correct?
I then solved it and got 1 as the answer.

If anybody could help me out that would be great.

Thx

Note that
$x_i=a+i\Delta x =1+\frac{i}{n}$

$\lim_{n \to \infty}\sum_{i=1}^{n}\left( 2+3(1+\frac{i}{n})^2\right)\frac{1}{n}= \lim_{n \to \infty}\left( \frac{2}{n}\sum_{i=1}^{n}1+\frac{3}{n}\sum_{i=1}^{ n}\left( 1+\frac{2i}{n}+\frac{i^2}{n^2}\right)\right)$

You should be able to finish from here

Edit Dang I type slow
• May 18th 2008, 10:09 PM
rexexdesign
Thank you both for the quick answer. I guess I got messed up with xi being 1+i/n

Thanks again (Rock)