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tukajo

  • 2 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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  1. tukajo
    • 2 years ago
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    @experimentX

  2. tukajo
    • 2 years ago
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    @Chlorophyll

  3. experimentX
    • 2 years ago
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    0 to 0 this should be 0

  4. tukajo
    • 2 years ago
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    Nope lol

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

  6. tukajo
    • 2 years ago
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    rectangle not square

  7. tukajo
    • 2 years ago
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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.

  8. experimentX
    • 2 years ago
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    |dw:1353065276769:dw|

  9. tukajo
    • 2 years ago
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    bottom one.

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

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

  12. tukajo
    • 2 years ago
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    book says this:|dw:1353065522208:dw|

  13. tukajo
    • 2 years ago
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    sorry for the pelletty drawing lol

  14. experimentX
    • 2 years ago
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    yeah you can do that!!

  15. tukajo
    • 2 years ago
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    But how do I write it out?

  16. tukajo
    • 2 years ago
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    Like the first step.. I guess..

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

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

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

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

  21. experimentX
    • 2 years ago
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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.

  22. tukajo
    • 2 years ago
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    but you are taking a limit as n approaches infinity for an exact answer...

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

  24. experimentX
    • 2 years ago
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    this is numerical integration, just replace x and y by x_i and y_i in f(x,y) \[ x_i = {i \over N}, y_i = {i \over M} \]

  25. tukajo
    • 2 years ago
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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?

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

  27. experimentX
    • 2 years ago
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    hold on!!

  28. tukajo
    • 2 years ago
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    haha sorry

  29. experimentX
    • 2 years ago
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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

  30. experimentX
    • 2 years ago
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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}]

  31. sirm3d
    • 2 years ago
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    \[\huge \Delta x = \frac{ 4 }{ N }, \Delta y = \frac{2}{N}\]

  32. experimentX
    • 2 years ago
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    write it as \[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 \]

  33. experimentX
    • 2 years ago
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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.

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

  35. experimentX
    • 2 years ago
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    must be something similar ... just fix your integration.

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

  37. experimentX
    • 2 years ago
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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

  38. experimentX
    • 2 years ago
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    http://tinyurl.com/bywugf4

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

  40. experimentX
    • 2 years ago
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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.

  41. Zarkon
    • 2 years ago
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    what a mess

  42. Zarkon
    • 2 years ago
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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 \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[3\left(i\frac{4}{n}\right)^2-2\left(j\frac{2}{n}\right)-2\right]\frac{4}{n}\frac{2}{n}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[48\frac{i^2}{n^2}-4\frac{j}{n}-2\right]\frac{8}{n^2}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[384\frac{i^2}{n^4}-32\frac{j}{n^3}-\frac{16}{n^2}\right]\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\left[384\frac{i^2}{n^4}n-32\frac{1}{n^3}\frac{n(n+1)}{2}-\frac{16}{n^2}n\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^4}n\frac{n(n+1)(2n+1)}{6}-32\frac{1}{n^3}\frac{n(n+1)}{2}n-\frac{16}{n^2}nn\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^2}\frac{(n+1)(2n+1)}{6}-32\frac{1}{n}\frac{(n+1)}{2}-16\right]\] \[=\frac{384\cdot 2}{6}-\frac{32}{2}-16=96\]

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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