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twirlere Group Title

y=x^2+1, [0,3] Use the limit process to find the area of the region between the graph of the function and the x-axis over the given interval.

  • one year ago
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  1. abb0t Group Title
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    First step is to find to find the right end points, ∆x, which is defined as: \(\Delta x = \frac{ b-a }{ n }\)

    • one year ago
  2. twirlere Group Title
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    3/n

    • one year ago
  3. abb0t Group Title
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    I'll leave it to you to verify your subintervals,, but as you can see,, the right endpoint on the \(i^{th}\) interval is going to be: \(\frac{ 3i }{ n }\).

    • one year ago
  4. abb0t Group Title
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    And the summation in the definition of the definite integral is defined as: \[\sum_{i=1}^{n}f(x_i)\Delta x\]

    • one year ago
  5. abb0t Group Title
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    In your case, you have: \[\sum_{i=1}^{n}f \left( \frac{ 3i }{ n } \right)\left( \frac{ 3 }{ n } \right)\]

    • one year ago
  6. abb0t Group Title
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    Are you with me so far? You're plugging in your right end point for "x". Hence: \[\sum_{i=1}^{n}\left[ \left( \frac{ 2i }{ n } \right)^2+1 \right]\left( \frac{ 2 }{ n } \right)\] Remember: \(f(x_i)\Delta x\)

    • one year ago
  7. twirlere Group Title
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    okay, I'm understanding...

    • one year ago
  8. abb0t Group Title
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    The rest is easy algebra that you should be able to do. Combine like terms and distribute. I'll leave that to you to work on.

    • one year ago
  9. abb0t Group Title
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    Now, remember that the definition is defined as: \[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}f(x_i)\Delta x\] So you're going to be taking the limit of this!

    • one year ago
  10. abb0t Group Title
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    In other words, you're going to have to use some of the formulas you used in algebra of summation notation to eliminate the actual summation and get a formular in terms of just "n". Because that's what you're taking the limit. You cannot have 'i' in the equation. Also, NOTE: "n" is a constant in this care. The only thing that changes in this is "i". So you can factor OUT "n".

    • one year ago
  11. abb0t Group Title
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    Here are the general formulas that you SHOULD know (sidenote: these are only true if i=1) \[\sum_{n}^{i=1}c = cn\] \[\sum_{i=1}^{n}i=\frac{ n(n+1) }{ 2 }\] \[\sum_{i=1}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }\] \[\sum_{i=1}^{n}i^3=\left[ \frac{ n(n+1) }{ 2 } \right]^2\]

    • one year ago
  12. abb0t Group Title
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    Are you still with me?

    • one year ago
  13. twirlere Group Title
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    yes

    • one year ago
  14. abb0t Group Title
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    I don't want to do all of yor work, especially not the algebra portion. So I'm going to ask, what did you get as your fuction? The one earlier: \(f(x_i) \Delta x\)? After combining the fraction into one single and distributing :)

    • one year ago
  15. twirlere Group Title
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    ((9i^2/n^2)+1)(3/n)

    • one year ago
  16. abb0t Group Title
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    I hope you used \[[(\frac{ 3i }{ n })^2+1](\frac{ 3 }{ n })\] because i i mistypoed a 2

    • one year ago
  17. satellite73 Group Title
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    really makes you appreciate the fundamental theorem of calculus, doesn't it?

    • one year ago
  18. twirlere Group Title
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    that's what I used... and no :( lol

    • one year ago
  19. satellite73 Group Title
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    no? compare with \[F(x)=\frac{x^3}{3}+x\] and then \(F(3)-F(0)\)

    • one year ago
  20. abb0t Group Title
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    I think that yourr result might be a tad bit off: \(\left[\frac{ 9i^2 }{ n^2 } +1\right] \frac{ 3 }{ n }\)

    • one year ago
  21. abb0t Group Title
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    Oh wait, nevermid, you were right. Now distribute. Haha.

    • one year ago
  22. abb0t Group Title
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    Also, remember the sum rules from I think algebra or pre-calculus? Where you can separate the sum: \[\sum_{i=1}^{n}(a \pm b) = \sum_{i=1}^{n}a \pm \sum_{i=1}^{n}b\]

    • one year ago
  23. twirlere Group Title
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    (27i^2/n^3)+(3/n)

    • one year ago
  24. abb0t Group Title
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    Great. So now, you can separete them! Good job. Now, remember what I told you, how n is a constant, and the only thing changing is "i"? Well, You can actually factor out the constant, n, in this case. Use the properties above:\[\frac{ 27 }{ n^3 } \sum_{i=1}^{n}i^2 + \frac{ 3 }{ n } \sum_{i=1}^{n}\] Use the properties i provided above to eliminate "i" and put it in terms of "n". Since you have \(i^2\) you'll be using the third formula, and for the second, use the first formula.

    • one year ago
  25. abb0t Group Title
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    Changing them you will get: \(\frac{ 27 }{ n^3 }\left( \frac{ n(n+1)(2n+1) }{ 6 } \right) + \frac{ 3 }{ n }(n)\). Can you do that? Cross cancel out the terms?

    • one year ago
  26. twirlere Group Title
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    \[(9(n+1)(2n+1)/2n^2) + 3\]

    • one year ago
  27. abb0t Group Title
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    Good job! Now, I think it might be easier to take the limit of this function if you combined it all into one (at least for me it's easier to see what cancels out). But all that's left to do is simply take the limit of that function! You're final result is your answer.

    • one year ago
  28. abb0t Group Title
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    Combined it all into one fraction*

    • one year ago
  29. twirlere Group Title
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    umm... I have trouble finding the limit

    • one year ago
  30. abb0t Group Title
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    Well, first off, did we get the same fraction, I did it fairly quick so I could be wrong so let's check that: \(\frac{ 21n^2+27n+9 }{ 2n^2 }\) are we on the same page?

    • one year ago
  31. amistre64 Group Title
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    \[\Large \frac{27}{n^3}~{\frac{n(n+1)(2n+1)}{6}}+3\] \[\Large {\frac{9n(n+1)(2n+1)+6n^3}{2n^3}}\] \[\Large {\frac{9n(2n^2+3n+1)+6n^3}{2n^3}}\] \[\Large {\frac{9(2+\cancel{\frac 3n+\frac1n)}^0+6}{2}}\] \[\Large 9+3\]

    • one year ago
  32. amistre64 Group Title
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    unless a typed it wrong of course .... latex is nice but consumes alot of brain power

    • one year ago
  33. abb0t Group Title
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    @amistre64 how did you do the "slash" across? :P

    • one year ago
  34. amistre64 Group Title
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    lol, ``` \cancel{arg} ```

    • one year ago
  35. amistre64 Group Title
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    and i put an ^0 on it to indicate it canceld

    • one year ago
  36. amistre64 Group Title
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    theres also an alternative cancel that goes against the other diagonal, /bcancel i think ... never ued it tho :)

    • one year ago
  37. abb0t Group Title
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    awe snaps!! this just got real!! Thanks :)

    • one year ago
  38. abb0t Group Title
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    Do you have a latex site for these codes? I have one but it's so basic. Lol.

    • one year ago
  39. amistre64 Group Title
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    whenever a see a neat latex written up on here, i right click it and "show math as tex commands" thatll copy the coding without the wrappers to inspect

    • one year ago
  40. amistre64 Group Title
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    try the Latex Practicing subject under "find more subjects" the math code they use here is not the complete setup, but it is pretty impressive nonetheless

    • one year ago
  41. abb0t Group Title
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    NEAT!

    • one year ago
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