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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)
 one year ago
 one year 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)
 one year ago
 one year ago

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experimentXBest ResponseYou've already chosen the best response.1
0 to 0 this should be 0
 one year ago

tukajoBest 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
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
so 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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1353065276769:dw
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
Now how do I write those as limits of riemann sums?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
do the usual way ... first with respect to dx dw:1353065464534:dw
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
book says this:dw:1353065522208:dw
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
sorry for the pelletty drawing lol
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah you can do that!!
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
But how do I write it out?
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
Like the first step.. I guess..
 one year ago

experimentXBest 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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[\Delta x = {1 \over N} \text{ and } \Delta y = {1\over M}\]
 one year ago

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

tukajoBest ResponseYou've already chosen the best response.0
Says use M = N for simplicity, where do I go from there?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
don'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.
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
but you are taking a limit as n approaches infinity for an exact answer...
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
and what do I put in for f(xij,yij)?
 one year ago

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

tukajoBest ResponseYou've already chosen the best response.0
The 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?
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
http://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}]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
\[\huge \Delta x = \frac{ 4 }{ N }, \Delta y = \frac{2}{N}\]
 one year ago

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

experimentXBest 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.
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
must be something similar ... just fix your integration.
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
wolf 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
 one year ago

tukajoBest ResponseYou've already chosen the best response.0
I think I'll ask my professor tomorrow. Sorry, I'm a bit lost. I appreciate the help.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
i,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.
 one year ago

ZarkonBest 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 \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[3\left(i\frac{4}{n}\right)^22\left(j\frac{2}{n}\right)2\right]\frac{4}{n}\frac{2}{n}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[48\frac{i^2}{n^2}4\frac{j}{n}2\right]\frac{8}{n^2}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[384\frac{i^2}{n^4}32\frac{j}{n^3}\frac{16}{n^2}\right]\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\left[384\frac{i^2}{n^4}n32\frac{1}{n^3}\frac{n(n+1)}{2}\frac{16}{n^2}n\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^4}n\frac{n(n+1)(2n+1)}{6}32\frac{1}{n^3}\frac{n(n+1)}{2}n\frac{16}{n^2}nn\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^2}\frac{(n+1)(2n+1)}{6}32\frac{1}{n}\frac{(n+1)}{2}16\right]\] \[=\frac{384\cdot 2}{6}\frac{32}{2}16=96\]
 one year ago
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