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nathanruff

Find the exact area bounded by f(x)=x^2 and the x-axis between x=1 and x=2 using n partitions.

  • one year ago
  • one year ago

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  1. ByteMe
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    n partitions imply a finite number... if you're going to find the EXACT area, you'll need an infinite number of very tiny partitions... ie, let \(\large n\rightarrow \infty \)

    • one year ago
  2. nathanruff
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    I ended up with this: \[A = (\frac{ 1 }{ n } + \frac{ 2(1) }{ n ^{2} } + \frac{ 1^{2} }{ n ^{3} }) + (\frac{ 1 }{ n } + \frac{ 2(2) }{ n ^{2} } + \frac{ 2^{2} }{ n ^{3} }) + (\frac{ 1 }{ n } + \frac{ 2(3) }{ n ^{2} } + \frac{ 3^{2} }{ n ^{3} }) + (\frac{ 1 }{ n } + \frac{ 2(4) }{ n ^{2} } + \frac{ 4^{2} }{ n ^{3} }) +...+ (\frac{ 1 }{ n } + \frac{ 2(n) }{ n ^{2} } + \frac{ n^{2} }{ n ^{3} })\] Simplified: \[A = \frac{ n }{ n } + \frac{ 2 }{ n ^{2} }(1+2+3+4+...+n)+\frac{ 1 }{ n ^{3} }(1^{2}+2^{2}+3^{2}+4^{2}+...+n ^{2})\] Is that correct? I know I have to take the limit of it eventually, but I know it simplifies more. That's what I'm unsure about.

    • one year ago
  3. ByteMe
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    |dw:1353405044723:dw|

    • one year ago
  4. nathanruff
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    Okay. I understand that.

    • one year ago
  5. ByteMe
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    \(\large f(1+\frac{i-1}{n})=(1+\frac{i-1}{n})^2 \) so the area of each rectangle is: \(\large A_i=(1+\frac{i-1}{n})^2 \cdot \frac{1}{n} \)

    • one year ago
  6. nathanruff
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    That matches with what I have. I'm just not sure if what I did next was right. Would you simplify that, or expand it (and distribute)?

    • one year ago
  7. ByteMe
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    yeah... sorry, i'm more visual so i had to draw that picture for me to see what i was doing...

    • one year ago
  8. ByteMe
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    yes.. expand that an add up all the \(\large A_i \) from 1 to n then take the limit as n approaches infinity...

    • one year ago
  9. ByteMe
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    yuck... looks like you'll have to expand \(\large (n + i - 1)^2 \)

    • one year ago
  10. ByteMe
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    lol... with a little help... http://www.wolframalpha.com/input/?i=expand+%28n%2Bi-1%29%5E2

    • one year ago
  11. ByteMe
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    whoa.. wait a minute... wolfram thinks i is the imaginary unit....

    • one year ago
  12. ByteMe
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    this is better... the index "i" is the "x" here: http://www.wolframalpha.com/input/?i=expand+%28n%2Bx-1%29%5E2

    • one year ago
  13. ByteMe
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    \(\large \lim_{n\rightarrow \infty} \sum_{i=1}^{n}A_i \)

    • one year ago
  14. nathanruff
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    I sort of understand what you're doing. The example in my textbook (Calculus 7E, Stewart) uses the formula for the sum of squares in simplifying this when taking my approach: \[A= \frac{ n }{ n } + \frac{ 2 }{ n ^{2} } (1+2+3+4+...+n) + \frac{ 1 }{ n ^{3} }(1^{2}+2^{2}+3^{2}+4^{2}+...+n ^{2})\] Where.. \[(1^{2}+2^{2}+3^{2}+4^{2}+...+n ^{2}) = \frac{ n(n+1)(2n+1) }{ 6 }\] Is there a way to simplify this? \[(1+2+3+4+...+n)\]

    • one year ago
  15. ByteMe
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    yes... that's n(n+1)/2

    • one year ago
  16. nathanruff
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    How did you do that? My textbook doesn't explain anything.

    • one year ago
  17. ByteMe
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    let the sum from 1 to n be x, and also, write it backwards, it's still x... |dw:1353407364634:dw|

    • one year ago
  18. ByteMe
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    now add theses two...|dw:1353407524440:dw|

    • one year ago
  19. ByteMe
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    so x, the sum, is n(n+1)/2.....

    • one year ago
  20. ByteMe
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    Stewart Calculus 7E ??? is this for AP calculus?

    • one year ago
  21. nathanruff
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    I'm taking Calculus I at a community college and my professor is supposedly the toughest calc instructor on campus. We can't use our calculators on any of the tests, there's no margin for error with him, and we have to memorize identities, theorems, and definitions.

    • one year ago