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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)

Mathematics
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Other answers:

Nope lol
[0,4]x[0,2] means a square from x = 0 to 4 and y = 0 to 2
rectangle not square
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.
|dw:1353065276769:dw|
bottom one.
Now how do I write those as limits of riemann sums?
do the usual way ... first with respect to dx |dw:1353065464534:dw|
book says this:|dw:1353065522208:dw|
sorry for the pelletty drawing lol
yeah you can do that!!
But how do I write it out?
Like the first step.. I guess..
\[x_i = {i \over N}, y_i = {i \over M}\] where N and M are your horizontal and vertical steps.
\[\Delta x = {1 \over N} \text{ and } \Delta y = {1\over M}\]
\[\lim n ->\infty of \sum_{i = 1}^{n} \sum_{j = 1}^{n} f(x _{i^*j},y_{ij^*})\Delta A\]
Says use M = N for simplicity, where do I go from there?
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.
but you are taking a limit as n approaches infinity for an exact answer...
and what do I put in for f(xij,yij)?
this is numerical integration, just replace x and y by x_i and y_i in f(x,y) \[ x_i = {i \over N}, y_i = {i \over M} \]
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?
\[ \sum_{i=0}^{N}\sum_{j=0}^{N} (3(x_i)^2 - 2y_i - 2)\; \Delta x \Delta y\]
hold on!!
haha sorry
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
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}]
\[\huge \Delta x = \frac{ 4 }{ N }, \Delta y = \frac{2}{N}\]
write it as \[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 \]
\[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 = \int_0^2 \int_0^4 (3x^2 - y - 2) dx dy\] I am not sure about your region.
j=1 and i = 1, not 0... so (3(1/n)^2 -2(1/n) -2)*(1/n)^2
must be something similar ... just fix your integration.
well I mean if you plugged in 0 for i and j wouldn't it erase the values?
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
http://tinyurl.com/bywugf4
I think I'll ask my professor tomorrow. Sorry, I'm a bit lost. I appreciate the help.
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.
what a mess
\[\lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{m}f(x_{ij}^*,y_{ij}^*)\Delta A\] 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 \(\displaystyle \Delta x=\frac{4-0}{n}=\frac{4}{n}\) and \(\displaystyle \Delta y=\frac{2-0}{n}=\frac{2}{n}\) \(\displaystyle x_{i}^*=i\frac{4}{n}\) and \(\displaystyle y_{j}^*=j\frac{2}{n}\) with \(f(x,y)=3x^2-2y-2\) this gives us \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[3\left(i\frac{4}{n}\right)^2-2\left(j\frac{2}{n}\right)-2\right]\frac{4}{n}\frac{2}{n}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[48\frac{i^2}{n^2}-4\frac{j}{n}-2\right]\frac{8}{n^2}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[384\frac{i^2}{n^4}-32\frac{j}{n^3}-\frac{16}{n^2}\right]\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\left[384\frac{i^2}{n^4}n-32\frac{1}{n^3}\frac{n(n+1)}{2}-\frac{16}{n^2}n\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^4}n\frac{n(n+1)(2n+1)}{6}-32\frac{1}{n^3}\frac{n(n+1)}{2}n-\frac{16}{n^2}nn\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^2}\frac{(n+1)(2n+1)}{6}-32\frac{1}{n}\frac{(n+1)}{2}-16\right]\] \[=\frac{384\cdot 2}{6}-\frac{32}{2}-16=96\]

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