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

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ByteMe
 3 years ago
Best ResponseYou've already chosen the best response.1n 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 \(\large n\rightarrow \infty \)

nathanruff
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

nathanruff
 3 years ago
Best ResponseYou've already chosen the best response.0Okay. I understand that.

ByteMe
 3 years ago
Best 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: \(\large A_i=(1+\frac{i1}{n})^2 \cdot \frac{1}{n} \)

nathanruff
 3 years ago
Best ResponseYou've already chosen the best response.0That 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)?

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

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

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

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

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

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

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

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

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

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

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

ByteMe
 3 years ago
Best ResponseYou've already chosen the best response.1so x, the sum, is n(n+1)/2.....

ByteMe
 3 years ago
Best ResponseYou've already chosen the best response.1Stewart Calculus 7E ??? is this for AP calculus?

nathanruff
 3 years ago
Best ResponseYou've already chosen the best response.0I'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.

ByteMe
 3 years ago
Best ResponseYou've already chosen the best response.1sorry man.... I have to take off... btw... that's what i used for AP calculus last year.... got a 1... :(