Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Kainui

  • one year ago

Is this a legitimate way to calculate integrals:

  • This Question is Closed
  1. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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: \[\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4-x^4-6x^2dx^2-4xdx^3-dx^4]\] continuing...

  2. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[\frac{ 15 }{ 4 } \int\limits_{a}^{b} [(x+dx)^4-x^4]-\frac{ 15 }{ 4 } dx \int\limits_{a}^{b}[6x^2dx+4xdx^2+dx^3]\] 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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  4. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  5. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  8. RolyPoly
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Riemann sum?!

  9. ybarrap
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  10. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    "delta n"

  12. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  13. ybarrap
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  14. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.