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tukajo

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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  1. tukajo
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    @experimentX

    • one year ago
  2. tukajo
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    @Chlorophyll

    • one year ago
  3. experimentX
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    0 to 0 this should be 0

    • one year ago
  4. tukajo
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    Nope lol

    • one year ago
  5. tukajo
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    [0,4]x[0,2] means a square from x = 0 to 4 and y = 0 to 2

    • one year ago
  6. tukajo
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    rectangle not square

    • one year ago
  7. tukajo
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    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
  8. experimentX
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    |dw:1353065276769:dw|

    • one year ago
  9. tukajo
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    bottom one.

    • one year ago
  10. tukajo
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    Now how do I write those as limits of riemann sums?

    • one year ago
  11. experimentX
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    do the usual way ... first with respect to dx |dw:1353065464534:dw|

    • one year ago
  12. tukajo
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    book says this:|dw:1353065522208:dw|

    • one year ago
  13. tukajo
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    sorry for the pelletty drawing lol

    • one year ago
  14. experimentX
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    yeah you can do that!!

    • one year ago
  15. tukajo
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    But how do I write it out?

    • one year ago
  16. tukajo
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    Like the first step.. I guess..

    • one year ago
  17. experimentX
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    \[x_i = {i \over N}, y_i = {i \over M}\] where N and M are your horizontal and vertical steps.

    • one year ago
  18. experimentX
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    \[\Delta x = {1 \over N} \text{ and } \Delta y = {1\over M}\]

    • one year ago
  19. tukajo
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    \[\lim n ->\infty of \sum_{i = 1}^{n} \sum_{j = 1}^{n} f(x _{i^*j},y_{ij^*})\Delta A\]

    • one year ago
  20. tukajo
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    Says use M = N for simplicity, where do I go from there?

    • one year ago
  21. experimentX
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    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
  22. tukajo
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    but you are taking a limit as n approaches infinity for an exact answer...

    • one year ago
  23. tukajo
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    and what do I put in for f(xij,yij)?

    • one year ago
  24. experimentX
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    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
  25. tukajo
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    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
  26. experimentX
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    \[ \sum_{i=0}^{N}\sum_{j=0}^{N} (3(x_i)^2 - 2y_i - 2)\; \Delta x \Delta y\]

    • one year ago
  27. experimentX
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    hold on!!

    • one year ago
  28. tukajo
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    haha sorry

    • one year ago
  29. experimentX
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    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
  30. experimentX
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    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
  31. sirm3d
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    \[\huge \Delta x = \frac{ 4 }{ N }, \Delta y = \frac{2}{N}\]

    • one year ago
  32. experimentX
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    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
  33. experimentX
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    \[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 = \int_0^2 \int_0^4 (3x^2 - y - 2) dx dy\] I am not sure about your region.

    • one year ago
  34. tukajo
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    j=1 and i = 1, not 0... so (3(1/n)^2 -2(1/n) -2)*(1/n)^2

    • one year ago
  35. experimentX
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    must be something similar ... just fix your integration.

    • one year ago
  36. tukajo
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    well I mean if you plugged in 0 for i and j wouldn't it erase the values?

    • one year ago
  37. experimentX
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    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
  38. experimentX
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    http://tinyurl.com/bywugf4

    • one year ago
  39. tukajo
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    I think I'll ask my professor tomorrow. Sorry, I'm a bit lost. I appreciate the help.

    • one year ago
  40. experimentX
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    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
  41. Zarkon
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    what a mess

    • one year ago
  42. Zarkon
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    \[\lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{m}f(x_{ij}^*,y_{ij}^*)\Delta A\] 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 \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[3\left(i\frac{4}{n}\right)^2-2\left(j\frac{2}{n}\right)-2\right]\frac{4}{n}\frac{2}{n}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[48\frac{i^2}{n^2}-4\frac{j}{n}-2\right]\frac{8}{n^2}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[384\frac{i^2}{n^4}-32\frac{j}{n^3}-\frac{16}{n^2}\right]\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\left[384\frac{i^2}{n^4}n-32\frac{1}{n^3}\frac{n(n+1)}{2}-\frac{16}{n^2}n\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^4}n\frac{n(n+1)(2n+1)}{6}-32\frac{1}{n^3}\frac{n(n+1)}{2}n-\frac{16}{n^2}nn\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^2}\frac{(n+1)(2n+1)}{6}-32\frac{1}{n}\frac{(n+1)}{2}-16\right]\] \[=\frac{384\cdot 2}{6}-\frac{32}{2}-16=96\]

    • one year ago
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