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

Calculus1
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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 \(\large n\rightarrow \infty \)
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.
|dw:1353405044723:dw|

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Other answers:

Okay. I understand that.
\(\large f(1+\frac{i-1}{n})=(1+\frac{i-1}{n})^2 \) so the area of each rectangle is: \(\large A_i=(1+\frac{i-1}{n})^2 \cdot \frac{1}{n} \)
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)?
yeah... sorry, i'm more visual so i had to draw that picture for me to see what i was doing...
yes.. expand that an add up all the \(\large A_i \) from 1 to n then take the limit as n approaches infinity...
yuck... looks like you'll have to expand \(\large (n + i - 1)^2 \)
lol... with a little help... http://www.wolframalpha.com/input/?i=expand+%28n%2Bi-1%29%5E2
whoa.. wait a minute... wolfram thinks i is the imaginary unit....
this is better... the index "i" is the "x" here: http://www.wolframalpha.com/input/?i=expand+%28n%2Bx-1%29%5E2
\(\large \lim_{n\rightarrow \infty} \sum_{i=1}^{n}A_i \)
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? \[(1+2+3+4+...+n)\]
yes... that's n(n+1)/2
How did you do that? My textbook doesn't explain anything.
let the sum from 1 to n be x, and also, write it backwards, it's still x... |dw:1353407364634:dw|
now add theses two...|dw:1353407524440:dw|
so x, the sum, is n(n+1)/2.....
Stewart Calculus 7E ??? is this for AP calculus?
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.
sorry man.... I have to take off... btw... that's what i used for AP calculus last year.... got a 1... :(
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!
yw....:)

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