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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.
 one year ago
 one year 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.
 one year ago
 one year 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 \(\large n\rightarrow \infty \)
 one year 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.
 one year ago

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

nathanruff Group TitleBest ResponseYou've already chosen the best response.0
Okay. I understand that.
 one year 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: \(\large A_i=(1+\frac{i1}{n})^2 \cdot \frac{1}{n} \)
 one year 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)?
 one year 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...
 one year 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...
 one year ago

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

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

ByteMe Group TitleBest ResponseYou've already chosen the best response.1
\(\large \lim_{n\rightarrow \infty} \sum_{i=1}^{n}A_i \)
 one year 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? \[(1+2+3+4+...+n)\]
 one year ago

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

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

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