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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?

  21. nathanruff
    • 2 years ago
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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.

  22. ByteMe
    • 2 years ago
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