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Summation of rectangular areas (calculus) problem. |
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| Oct9-04, 11:14 PM | #1 |
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Summation of rectangular areas (calculus) problem.
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 [tex]= \sum_{k=1}^n\ f(x) * (3/n)[/tex] [tex]= \sum_{k=1}^n\ [(k * 3/n)^2 + 2(k * 3/n)] * (3/n)[/tex] [tex]= \sum_{k=1}^n\ [k^2 * (3/n)^2 + 2 * k * (3/n)] * (3/n)[/tex] [tex]= \sum_{k=1}^n\ [k^2 * (3/n) + 2k] * (3/n)^2[/tex] [tex]= 9/n^2\sum_{k=1}^n\ [k^2 * (3/n) + 2k)][/tex] 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 [tex]\sum_{k=1}^n\ k^2[/tex] would become n(n+1)(2n + 1) / 6 and I could go straight to the limits |
| Oct9-04, 11:41 PM | #2 |
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I'm not sure what your difficulty is - it appears you're right on the mark!
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| Oct10-04, 12:06 AM | #3 |
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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! |
| Oct10-04, 11:05 AM | #4 |
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Summation of rectangular areas (calculus) problem.
Could you experts tell me if what I'm doing below is "legal" in terms of math
[tex]9/n^2\sum_{k=1}^n\ k^2 * (3/n) + 2k[/tex] becomes [tex]= 9/n^2[\sum_{k=1}^n\ k^2 * (3/n) + \sum_{k=1}^n\ 2k][/tex] [tex]= 9/n^2 [(3/n)\sum_{k=1}^n\ k^2 + 2\sum_{k=1}^n\ k][/tex] 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 |
| Oct14-04, 01:29 AM | #5 |
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Recognitions:
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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. |
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