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Definite Integrals

  1. Dec 2, 2007 #1
    [SOLVED] Definite Integrals

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Why is the answer 26/3? I got 4 by using the limit/Riemann Sum definition. Is this one method to calculate definite integrals?
  2. jcsd
  3. Dec 2, 2007 #2
    by just taking the anti-derivative, i get 26/3. check your work again.
    Last edited: Dec 2, 2007
  4. Dec 2, 2007 #3
    Sorry, but I don't think I can use any other methods because of my teacher.

    I got 2n for the width of the rectangle and 1+(2i/n) for the height..

    AHHHH: I found my error.. I forgot to square the 2i/n -_____-
  5. Dec 2, 2007 #4
    Oh jesus christ.... My answer never matches. Perhaps I should do everything step-by-step?
  6. Dec 2, 2007 #5
    Thanks a lot for bothering to post those other methods. I hope I will get to learn/use them later!
  7. Dec 2, 2007 #6
    hey, show some work. But if u are stuck as where to start.. do these
    find Delta X, find Xi. and then use this formula

    lim [tex]\sum f(Xi)\Delta X[/tex]
    n [tex] \rightarrow \infty [/tex]

    ugh thats so bad... i have never had to use this.. but like..i'm sure u can find it on wikipedia
    In addition, know what the summation of a X^2 series is.. it is n(n+1)(2n+1)6\

    ah, i feel so bad.. i couldn't draw it successfully.
    Last edited: Dec 2, 2007
  8. Dec 2, 2007 #7
    so basically it is
    2/n[tex]\sum (1+2i/n)^2[/tex]

    then use foil(or whatever u call it).

    it becomes (1+4i/n +4i^2/n)
    Last edited: Dec 2, 2007
  9. Dec 2, 2007 #8
    Yeah, thanks aq1q. That's what I did but I didn't square the x term for the height. I also messed up on simplifying. I just make too many mistakes :/
  10. Dec 2, 2007 #9
    ah :\ so everything ok now?
  11. Dec 2, 2007 #10

    Take the integral:

    [(1/3)(x)^3] over 3 and 1

    Then evaluate:

    [(1/3)(3)^3] - [(1/3)(1)^3] = 26/3
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