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Kainui
 one year ago
Best ResponseYou've already chosen the best response.4I'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...

Kainui
 one year ago
Best 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: \[15\frac{ b^4a^4 }{ 4 }=\int\limits_{a}^{b}15x^3dx\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.4Call me out on whatever seems the most shaky in my reasoning here if you want some explanation, cause I know it's pretty... weird.

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

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

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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.4Do 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.

ybarrap
 one year ago
Best 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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.4I 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?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.4I was fairly certain that summations only ever counted by 1's.

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0If 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.
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