# Summation of rectangular areas (calculus) problem.

1. Oct 9, 2004

### singleton

Good evening. I'm having a little difficulty with the summation of rectangular areas when finding the area under a curve.

Question:
Using summation of rectangles, find the area enclosed between the curve y = x^2 + 2x and the x-axis from x=0 to x=3.

Well, I start by dividing the interval (from x=0 to x=3) by n equal parts to find the width of each rectangular area.
=3/n

Then I begin using sigma notation (I'm new at this)
Sum of rectangular areas
$$= \sum_{k=1}^n\ f(x) * (3/n)$$
$$= \sum_{k=1}^n\ [(k * 3/n)^2 + 2(k * 3/n)] * (3/n)$$
$$= \sum_{k=1}^n\ [k^2 * (3/n)^2 + 2 * k * (3/n)] * (3/n)$$
$$= \sum_{k=1}^n\ [k^2 * (3/n) + 2k] * (3/n)^2$$
$$= 9/n^2\sum_{k=1}^n\ [k^2 * (3/n) + 2k)]$$

Now my main problem is that I'm trying to isolate the k^2 so that I can write out the summation formula for it and then go to limits and discover the area under the curve.

eg
$$\sum_{k=1}^n\ k^2$$
would become
n(n+1)(2n + 1) / 6
and I could go straight to the limits

Last edited: Oct 9, 2004
2. Oct 9, 2004

### Tide

I'm not sure what your difficulty is - it appears you're right on the mark!

3. Oct 10, 2004

### singleton

The difficulty is that I don't know -how- to isolate the k^2 :(

I know very little about using this sigma notation, the only rule I know is that when you have sigma(ak + b) you can rewrite as:
sigma(ak) + sigma(b)
=(a)sigma(k) + (b)sigma(1)

where a and b are constants.

for me, the problem is that there is a variable on the other side of the addition--2k that is...

Again, I know very little of the rules of sigma notation. Perhaps there is a way to rewrite this? I'm not wanting the answer for the area--rather just a more manageable way for me to put it to limits ;)

Or is it already to go??? I had just figured I could "do more" to it.

Thanks again!

4. Oct 10, 2004

### singleton

Could you experts tell me if what I'm doing below is "legal" in terms of math

$$9/n^2\sum_{k=1}^n\ k^2 * (3/n) + 2k$$
becomes
$$= 9/n^2[\sum_{k=1}^n\ k^2 * (3/n) + \sum_{k=1}^n\ 2k]$$
$$= 9/n^2 [(3/n)\sum_{k=1}^n\ k^2 + 2\sum_{k=1}^n\ k]$$

And then the first sigma notation would become
n(n + 1)(2n + 1)
--------------
6

and the second sigma notation would become
n(n + 1)
--------
2

and I could then expand the 9/n^2 on it and evaluate the limits? This is probably really stupid question but yesterday was the first day I worked with this notation :D

5. Oct 14, 2004

### Tide

Single,

Sorry for the late reply - I lost track of you there!

What you did is fine and you're basically done. Now just ask yourself what the limiting value of your expression is as you let n go to infinity.