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Kainui Group TitleBest ResponseYou've already chosen the best response.4
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: \[\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4x^46x^2dx^24xdx^3dx^4]\] continuing...
 11 months ago

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

Kainui Group TitleBest ResponseYou've already chosen the best response.4
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.
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
it seems to work for all polynomials... i got mixed feelings hmm but i don't see any flaw in the idea
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
u have used the differential as \(\Delta x\) and abused filly ! lol it looks good xD
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.0
I would change the notation: use summations , \(\Delta x\), and then take the limit as Δx >0
 11 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.4
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.
 11 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Riemann sum?!
 11 months ago

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
@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
 11 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.4
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?
 11 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.4
I was fairly certain that summations only ever counted by 1's.
 11 months ago

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
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.
 11 months ago
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