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twirlere

  • one year ago

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.

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

  2. twirlere
    • one year ago
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    3/n

  3. abb0t
    • one year ago
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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: \(\frac{ 3i }{ n }\).

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

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

  6. abb0t
    • one year ago
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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: \(f(x_i)\Delta x\)

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

  8. abb0t
    • one year ago
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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.

  9. abb0t
    • one year ago
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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!

  10. abb0t
    • one year ago
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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".

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

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

  13. twirlere
    • one year ago
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    yes

  14. abb0t
    • one year ago
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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 :)

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

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

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

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

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

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

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

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

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

  24. abb0t
    • one year ago
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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.

  25. abb0t
    • one year ago
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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?

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

  27. abb0t
    • one year ago
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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.

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

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

  30. abb0t
    • one year ago
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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?

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

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

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

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

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

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

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

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

  39. amistre64
    • one year ago
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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

  40. amistre64
    • one year ago
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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

  41. abb0t
    • one year ago
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    NEAT!

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