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VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: #1procrastinator on July 13, 2012, 09:37:37 am

Title: Prove this using Riemann sums
Post by: #1procrastinator on July 13, 2012, 09:37:37 am: Prove this:
$\int_a^b \! x \, \mathrm{d} x = \frac{b^2-a^2}{2}$
using Riemann sums

This is the formula I've got and according to my TI-89, it's the right formula because it evaluates to the above but I'm really struggling with the algebra. A hint would be much appreciated.

$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}[\frac{(b-a)i}{n}+a][\frac{b-a}{n}]$

Also, what's the difference between

$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}$

and

$\sum_{i=1}^{\infty}$

EDIT: not sure why tex isn't working cause it shows up fine in the sandbox
Title: Re: Prove this using Riemann sums
Post by: kamil9876 on July 13, 2012, 11:45:19 am: Quote from: #1procrastinator on July 13, 2012, 09:37:37 am
Prove this:
$\int_a^b \! x \, \mathrm{d} x = \frac{b^2-a^2}{2}$
using Riemann sums

(Technical note: I'm assuming that you are assuming that you know that the integral exists, and you just want to calculate it's value using Riemann Sums, i.e we don't care about proving that it is Riemann integrable directly from the definition... it's probably doable but more work obviously)

So let $I_n$ be the rectangular left hand approximation with n evenly spaced intervals.

$I_n=\sum_{k=0}^{n-1} \frac{b-a}{n} (a+k\frac{b-a}{n})$

$=\frac{b-a}{n} \sum_{k=0}^{n-1}(a + k\frac{b-a}{n})$

$=\frac{b-a}{n} (na + \frac{b-a}{n} \sum_{k=0}^{n-1} k )$

Now the sum should be easy to evaluate.

Here is an extension exercise you may want to try:

a) Let $m$ be a natural number, $S(n)=\sum_{k=1}^n k^m$ . It is well known that $S(n)$ is a polynomial in $n$ of degree $m+1$ with leading coefficient (i.e coefficient of $n^{m+1}$ ) equal to \frac{1}{m+1}. Use this to evaluate $\int_ a^b x^m dx$ for natural number m using Riemann sums. (a=0,b=1 is a good enough exercise)

b) Prove the claim given in part a (hint, $S(n+1)-S(n)=(n+1)^m$ , now expand and equate coefficients)

Quote from: #1procrastinator on July 13, 2012, 09:37:37 am

Also, what's the difference between

$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}$

and

$\sum_{i=1}^{\infty}$

The former is the definition of the latter (infinite sums don't really make sense in the reals, but limits do)
Title: Re: Prove this using Riemann sums
Post by: #1procrastinator on July 13, 2012, 02:23:30 pm: ^ Yeah, the question really just said to prove it. Didn't say which method to use but the whole chapter was on Riemann sums and there's not yet any mention of the antiderivative

I'm not sure how you got line 3 - which property of sums is that?

Also, I see that you don't put brackets around $\frac{b-a}{n}$ when you multiply by the integer...just trying to understand that intuitively right not (in no way doubting! :p)...and you didn't write the limit thing, is that implied somewhere?

By the way, is the expression I got possible to do my hand?

EDIT: For the last question, so do you mean that it doesn't make sense for it to sum everything to infinity but only as n gets super close to infinitely close infinity? lol
Title: Re: Prove this using Riemann sums
Post by: kamil9876 on July 13, 2012, 04:51:50 pm: I was calculating $I_n$ the nth approximation. The idea is so simplify that sum (use the arithmetic sequence formula) to get a nice expression for I_n. Then let n go to infinity and you get the result.

Quote
I see that you don't put brackets around \frac{b-a}{n} when you multiply by the integer.

Yes, laziness.

Quote
I'm not sure how you got line 3 - which property of sums is that?

Linearity: $\Sigma(x_i+y_i)=\Sigma(x_i)+\Sigma(y_i)$ and $\Sigma{ax_i}=a\Sigma{x_i}$ if a is a constant. Also if you sum a constant n times you get n times that constant.

Quote
EDIT: For the last question, so do you mean that it doesn't make sense for it to sum everything to infinity but only as n gets super close to infinitely close infinity? lol

what does $L=\lim_{n\to \infty} s_n$ mean? it means that for every $\epsilon>0$ there is a natural number $N$ such that if $n>N$ then $|s_n-L|<\epsilon$ .
Title: Re: Prove this using Riemann sums
Post by: #1procrastinator on July 16, 2012, 12:26:35 pm: Ah thanks a lot, I don't think I was aware of the second property