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anonymous
 3 years ago
f(x,y) = 3x^2  2y  2
Region R = [0,4]x[0,2]
Compute the integral exactly by using the definition in terms of a limit of a Riemann Sum. For convenience take m = n. (M = horizontal slices, N = Vertical slices, 2 of each)
anonymous
 3 years ago
f(x,y) = 3x^2  2y  2 Region R = [0,4]x[0,2] Compute the integral exactly by using the definition in terms of a limit of a Riemann Sum. For convenience take m = n. (M = horizontal slices, N = Vertical slices, 2 of each)

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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.10 to 0 this should be 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0[0,4]x[0,2] means a square from x = 0 to 4 and y = 0 to 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so we need to integrate with respect x and respect to y but I am not sure how to write those integrals as limits of a riemann sum.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1353065276769:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now how do I write those as limits of riemann sums?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1do the usual way ... first with respect to dx dw:1353065464534:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0book says this:dw:1353065522208:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry for the pelletty drawing lol

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yeah you can do that!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But how do I write it out?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Like the first step.. I guess..

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[x_i = {i \over N}, y_i = {i \over M}\] where N and M are your horizontal and vertical steps.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Delta x = {1 \over N} \text{ and } \Delta y = {1\over M}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim n >\infty of \sum_{i = 1}^{n} \sum_{j = 1}^{n} f(x _{i^*j},y_{ij^*})\Delta A\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Says use M = N for simplicity, where do I go from there?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1don't need to put n>inf \[\sum_{i=0}^{N}\sum_{j=0}^{N}f(x_i, y_)\; \Delta x \Delta y\] just put N = 10 or 20 ... or any value you like. greater value you put, more accurate you get.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but you are taking a limit as n approaches infinity for an exact answer...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and what do I put in for f(xij,yij)?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1this is numerical integration, just replace x and y by x_i and y_i in f(x,y) \[ x_i = {i \over N}, y_i = {i \over M} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The teacher had some weird like Lim n> infty of (64n^2 + 12n + 33)/n^2 at the end of his problem when he did it?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ \sum_{i=0}^{N}\sum_{j=0}^{N} (3(x_i)^2  2y_i  2)\; \Delta x \Delta y\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=Sum [Sum[%283%28i%2Fn%29^2++2%28j%2Fn%29++2%29%281%2Fn%29^2%2C+{i%2C+0%2C+2n}]%2C{j%2C+0%2C+4n}]%2F.n%3Einfinity

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=Integrate [Integrate[3+x^2++2+y++2%2C+{x%2C+0%2C+2}]%2C+{y%2C+0%2C+4}]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge \Delta x = \frac{ 4 }{ N }, \Delta y = \frac{2}{N}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1write it as \[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2  2\left(j \over n\right)  2\right)\left(1 \over n\right)^2 \]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2  2\left(j \over n\right)  2\right)\left(1 \over n\right)^2 = \int_0^2 \int_0^4 (3x^2  y  2) dx dy\] I am not sure about your region.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0j=1 and i = 1, not 0... so (3(1/n)^2 2(1/n) 2)*(1/n)^2

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1must be something similar ... just fix your integration.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well I mean if you plugged in 0 for i and j wouldn't it erase the values?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1wolf seems to get answer from i,j=0 http://www.wolframalpha.com/input/?i=Sum [Sum[%283%28i%2Fn%29^2++2%28j%2Fn%29++2%29%281%2Fn%29^2%2C+{i%2C+0%2C+2n}]%2C{j%2C+0%2C+4n}]%2F.n%3Einfinity

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I'll ask my professor tomorrow. Sorry, I'm a bit lost. I appreciate the help.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1i,j=0 would mean you are taking the lower Riemann sum. since you are taking n>inf, naturally for lower values of i,j, the functional values tend to zero ... so it shouldn't matter much. If you want to stick to defn then it's fine as well.

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{m}f(x_{ij}^*,y_{ij}^*)\Delta A\] if we let \(n=m\) and pick upper right endpoints we get \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}f(x_{i}^*,y_{j}^*)\Delta A\] Here \(\Delta A=\Delta x \Delta y\) where \(\displaystyle \Delta x=\frac{40}{n}=\frac{4}{n}\) and \(\displaystyle \Delta y=\frac{20}{n}=\frac{2}{n}\) \(\displaystyle x_{i}^*=i\frac{4}{n}\) and \(\displaystyle y_{j}^*=j\frac{2}{n}\) with \(f(x,y)=3x^22y2\) this gives us \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[3\left(i\frac{4}{n}\right)^22\left(j\frac{2}{n}\right)2\right]\frac{4}{n}\frac{2}{n}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[48\frac{i^2}{n^2}4\frac{j}{n}2\right]\frac{8}{n^2}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[384\frac{i^2}{n^4}32\frac{j}{n^3}\frac{16}{n^2}\right]\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\left[384\frac{i^2}{n^4}n32\frac{1}{n^3}\frac{n(n+1)}{2}\frac{16}{n^2}n\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^4}n\frac{n(n+1)(2n+1)}{6}32\frac{1}{n^3}\frac{n(n+1)}{2}n\frac{16}{n^2}nn\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^2}\frac{(n+1)(2n+1)}{6}32\frac{1}{n}\frac{(n+1)}{2}16\right]\] \[=\frac{384\cdot 2}{6}\frac{32}{2}16=96\]
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