anonymous
  • anonymous
Find the exact area bounded by f(x)=x^2 and the x-axis between x=1 and x=2 using n partitions.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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 \(\large n\rightarrow \infty \)
anonymous
  • anonymous
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.
anonymous
  • anonymous
|dw:1353405044723:dw|

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anonymous
  • anonymous
Okay. I understand that.
anonymous
  • anonymous
\(\large f(1+\frac{i-1}{n})=(1+\frac{i-1}{n})^2 \) so the area of each rectangle is: \(\large A_i=(1+\frac{i-1}{n})^2 \cdot \frac{1}{n} \)
anonymous
  • anonymous
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)?
anonymous
  • anonymous
yeah... sorry, i'm more visual so i had to draw that picture for me to see what i was doing...
anonymous
  • anonymous
yes.. expand that an add up all the \(\large A_i \) from 1 to n then take the limit as n approaches infinity...
anonymous
  • anonymous
yuck... looks like you'll have to expand \(\large (n + i - 1)^2 \)
anonymous
  • anonymous
lol... with a little help... http://www.wolframalpha.com/input/?i=expand+%28n%2Bi-1%29%5E2
anonymous
  • anonymous
whoa.. wait a minute... wolfram thinks i is the imaginary unit....
anonymous
  • anonymous
this is better... the index "i" is the "x" here: http://www.wolframalpha.com/input/?i=expand+%28n%2Bx-1%29%5E2
anonymous
  • anonymous
\(\large \lim_{n\rightarrow \infty} \sum_{i=1}^{n}A_i \)
anonymous
  • anonymous
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? \[(1+2+3+4+...+n)\]
anonymous
  • anonymous
yes... that's n(n+1)/2
anonymous
  • anonymous
How did you do that? My textbook doesn't explain anything.
anonymous
  • anonymous
let the sum from 1 to n be x, and also, write it backwards, it's still x... |dw:1353407364634:dw|
anonymous
  • anonymous
now add theses two...|dw:1353407524440:dw|
anonymous
  • anonymous
so x, the sum, is n(n+1)/2.....
anonymous
  • anonymous
Stewart Calculus 7E ??? is this for AP calculus?
anonymous
  • anonymous
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.
anonymous
  • anonymous
sorry man.... I have to take off... btw... that's what i used for AP calculus last year.... got a 1... :(
anonymous
  • anonymous
No worries. So far, the whole class is hanging by a thread grade-wise. Again, I really do appreciate the help! You're a lifesaver! Thank you ByteMe!
anonymous
  • anonymous
yw....:)

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