anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@experimentX
anonymous
  • anonymous
@Chlorophyll
experimentX
  • experimentX
0 to 0 this should be 0

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anonymous
  • anonymous
Nope lol
anonymous
  • anonymous
[0,4]x[0,2] means a square from x = 0 to 4 and y = 0 to 2
anonymous
  • anonymous
rectangle not square
anonymous
  • anonymous
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.
experimentX
  • experimentX
|dw:1353065276769:dw|
anonymous
  • anonymous
bottom one.
anonymous
  • anonymous
Now how do I write those as limits of riemann sums?
experimentX
  • experimentX
do the usual way ... first with respect to dx |dw:1353065464534:dw|
anonymous
  • anonymous
book says this:|dw:1353065522208:dw|
anonymous
  • anonymous
sorry for the pelletty drawing lol
experimentX
  • experimentX
yeah you can do that!!
anonymous
  • anonymous
But how do I write it out?
anonymous
  • anonymous
Like the first step.. I guess..
experimentX
  • experimentX
\[x_i = {i \over N}, y_i = {i \over M}\] where N and M are your horizontal and vertical steps.
experimentX
  • experimentX
\[\Delta x = {1 \over N} \text{ and } \Delta y = {1\over M}\]
anonymous
  • anonymous
\[\lim n ->\infty of \sum_{i = 1}^{n} \sum_{j = 1}^{n} f(x _{i^*j},y_{ij^*})\Delta A\]
anonymous
  • anonymous
Says use M = N for simplicity, where do I go from there?
experimentX
  • experimentX
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.
anonymous
  • anonymous
but you are taking a limit as n approaches infinity for an exact answer...
anonymous
  • anonymous
and what do I put in for f(xij,yij)?
experimentX
  • experimentX
this is numerical integration, just replace x and y by x_i and y_i in f(x,y) \[ x_i = {i \over N}, y_i = {i \over M} \]
anonymous
  • anonymous
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?
experimentX
  • experimentX
\[ \sum_{i=0}^{N}\sum_{j=0}^{N} (3(x_i)^2 - 2y_i - 2)\; \Delta x \Delta y\]
experimentX
  • experimentX
hold on!!
anonymous
  • anonymous
haha sorry
experimentX
  • experimentX
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
experimentX
  • experimentX
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}]
sirm3d
  • sirm3d
\[\huge \Delta x = \frac{ 4 }{ N }, \Delta y = \frac{2}{N}\]
experimentX
  • experimentX
write it as \[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 \]
experimentX
  • experimentX
\[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 = \int_0^2 \int_0^4 (3x^2 - y - 2) dx dy\] I am not sure about your region.
anonymous
  • anonymous
j=1 and i = 1, not 0... so (3(1/n)^2 -2(1/n) -2)*(1/n)^2
experimentX
  • experimentX
must be something similar ... just fix your integration.
anonymous
  • anonymous
well I mean if you plugged in 0 for i and j wouldn't it erase the values?
experimentX
  • experimentX
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
experimentX
  • experimentX
http://tinyurl.com/bywugf4
anonymous
  • anonymous
I think I'll ask my professor tomorrow. Sorry, I'm a bit lost. I appreciate the help.
experimentX
  • experimentX
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.
Zarkon
  • Zarkon
what a mess
Zarkon
  • Zarkon
\[\lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{m}f(x_{ij}^*,y_{ij}^*)\Delta A\] 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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