## Kainui Group Title Is this a legitimate way to calculate integrals: 9 months ago 9 months ago

1. Kainui

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...

2. Kainui

$\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4-x^4]-\frac{ 15 }{ 4 } dx \int\limits_{a}^{b}[6x^2dx+4xdx^2+dx^3]$ 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. Kainui

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.

4. ganeshie8

it seems to work for all polynomials... i got mixed feelings hmm but i don't see any flaw in the idea

5. ganeshie8

u have used the differential as $$\Delta x$$ and abused filly ! lol it looks good xD

6. phi

I would change the notation: use summations , $$\Delta x$$, and then take the limit as Δx ->0

7. Kainui

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.

8. RolyPoly

Riemann sum?!

9. ybarrap

@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

10. Kainui

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. Kainui

"delta n"

12. Kainui

I was fairly certain that summations only ever counted by 1's.

13. ybarrap

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.