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Kainui

Is this a legitimate way to calculate integrals:

  • 3 months ago
  • 3 months ago

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  1. Kainui
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    I'll just start with this arbitrary integral: \[\int\limits_{a}^{b}15x^3dx\] now I notice that: \[(x+dx)^4=x^4+4x^3dx+6x^2dx^2+4xdx^3+dx^4\] and then I solve for x^3dx in the middle there and plug it in: \[\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4-x^4-6x^2dx^2-4xdx^3-dx^4]\] continuing...

    • 3 months ago
  2. Kainui
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    \[\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4-x^4]-\frac{ 15 }{ 4 } dx \int\limits_{a}^{b}[6x^2dx+4xdx^2+dx^3]\] Since the whole right side is multiplied by dx, that's essentially the limit as it approaches zero, so it's gone. \[\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4-x^4]\] This is just a telescoping series: \[\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4-x^4] = \frac{ 15 }{ 4 }[(a+dx)^4-a^4]+\frac{ 15 }{ 4 }[(a+2dx)^4-(a+dx)^4]+...+\frac{ 15 }{ 4 }[(b+dx)^4-b^4]\] So this simplifies down to: \[\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4-x^4] = \frac{ 15 }{ 4 }[(b+dx)^4-a^4]\] and again we take the limit as dx approaches zero and get the correct answer: \[15\frac{ b^4-a^4 }{ 4 }=\int\limits_{a}^{b}15x^3dx\]

    • 3 months ago
  3. Kainui
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    Call me out on whatever seems the most shaky in my reasoning here if you want some explanation, cause I know it's pretty... weird.

    • 3 months ago
  4. ganeshie8
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    it seems to work for all polynomials... i got mixed feelings hmm but i don't see any flaw in the idea

    • 3 months ago
  5. ganeshie8
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    u have used the differential as \(\Delta x\) and abused filly ! lol it looks good xD

    • 3 months ago
  6. phi
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    I would change the notation: use summations , \(\Delta x\), and then take the limit as Δx ->0

    • 3 months ago
  7. Kainui
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    Do summations work like that though? For instance, if the summation is between a and b, then as delta x approaches zero, the summation will just approach zero as well.

    • 3 months ago
  8. RolyPoly
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    Riemann sum?!

    • 3 months ago
  9. ybarrap
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    @Kainui, by using sums and deltas, that's exactly what Riemann Sums are -- the sums of a finite number of rectangles with width \(\Delta x\) and height \(f(x)\). If the sums as \(\Delta x \to 0\) then the continuous integral exists, and you can use \(\int\): http://en.wikipedia.org/wiki/Riemann_sum

    • 3 months ago
  10. Kainui
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    I guess, I saw that once before. lol. So if delta x = 1/2 for this summation: \[\sum_{n=2}^{4}n=2+5/2+3+7/2+4\] Right?

    • 3 months ago
  11. Kainui
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    "delta n"

    • 3 months ago
  12. Kainui
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    I was fairly certain that summations only ever counted by 1's.

    • 3 months ago
  13. ybarrap
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    If your interval is [2,4] then \(\Delta x= \cfrac{2}{n}\), where n is the number of partitions of the interval. We will let n go to infinity and compute the Riemann Sum of areas: \(x^2\times\Delta x=(2+\cfrac{2k}{n})^2\times \cfrac{2}{n}\) as \(n\to \infty\) in this interval. Here \(k\) denotes the \(k^{th} \) partition of the n partitions in [2,4]: $$ \sum_{i=1}^{n}(2+\cfrac{2k}{n})^2\times \cfrac{2}{n} \\ =\cfrac{8}{n^3}\sum_{k=1}^{n}(n+k)^2\\ =\cfrac{8}{n^3}\sum_{k=1}^{n}\left ( n^2+2nk+k^2\right )\\ =8\sum_{k=1}^{n}\left ( \cfrac{1}{n}+\cfrac{2k}{n^2}+\cfrac{k^2}{n^3} \right )\\ =8\left [ 1+1+\cfrac{1}{3}\right ]\text{ as }n\to\infty \\ =\cfrac{56}{3} $$ This is exactly the integral of $$ \int_2^4x^2dx $$ Note that the Riemann sums here did not just count by 1's, but the indices (\(k\)) did.

    • 3 months ago
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