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nathanruff

  • 2 years ago

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

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  1. ByteMe
    • 2 years ago
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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 \(\large n\rightarrow \infty \)

  2. nathanruff
    • 2 years ago
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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.

  3. ByteMe
    • 2 years ago
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    |dw:1353405044723:dw|

  4. nathanruff
    • 2 years ago
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    Okay. I understand that.

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

  6. nathanruff
    • 2 years ago
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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)?

  7. ByteMe
    • 2 years ago
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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...

  8. ByteMe
    • 2 years ago
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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...

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

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

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

  12. ByteMe
    • 2 years ago
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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

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

  14. nathanruff
    • 2 years ago
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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? \[(1+2+3+4+...+n)\]

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

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

  17. ByteMe
    • 2 years ago
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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|

  18. ByteMe
    • 2 years ago
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    now add theses two...|dw:1353407524440:dw|

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

  20. ByteMe
    • 2 years ago
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    Stewart Calculus 7E ??? is this for AP calculus?