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nathanruff
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Find the exact area bounded by f(x)=x^2 and the xaxis between x=1 and x=2 using n partitions.
 2 years ago
 2 years ago
nathanruff Group Title
Find the exact area bounded by f(x)=x^2 and the xaxis between x=1 and x=2 using n partitions.
 2 years ago
 2 years ago

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ByteMe Group TitleBest ResponseYou've already chosen the best response.1
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 years ago

nathanruff Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
dw:1353405044723:dw
 2 years ago

nathanruff Group TitleBest ResponseYou've already chosen the best response.0
Okay. I understand that.
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
\(\large f(1+\frac{i1}{n})=(1+\frac{i1}{n})^2 \) so the area of each rectangle is: \(\large A_i=(1+\frac{i1}{n})^2 \cdot \frac{1}{n} \)
 2 years ago

nathanruff Group TitleBest ResponseYou've already chosen the best response.0
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)?
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
yeah... sorry, i'm more visual so i had to draw that picture for me to see what i was doing...
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
yes.. expand that an add up all the \(\large A_i \) from 1 to n then take the limit as n approaches infinity...
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
yuck... looks like you'll have to expand \(\large (n + i  1)^2 \)
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
lol... with a little help... http://www.wolframalpha.com/input/?i=expand+%28n%2Bi1%29%5E2
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
whoa.. wait a minute... wolfram thinks i is the imaginary unit....
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
this is better... the index "i" is the "x" here: http://www.wolframalpha.com/input/?i=expand+%28n%2Bx1%29%5E2
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
\(\large \lim_{n\rightarrow \infty} \sum_{i=1}^{n}A_i \)
 2 years ago

nathanruff Group TitleBest ResponseYou've already chosen the best response.0
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? \[(1+2+3+4+...+n)\]
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
yes... that's n(n+1)/2
 2 years ago

nathanruff Group TitleBest ResponseYou've already chosen the best response.0
How did you do that? My textbook doesn't explain anything.
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
let the sum from 1 to n be x, and also, write it backwards, it's still x... dw:1353407364634:dw
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
now add theses two...dw:1353407524440:dw
 2 years ago