tukajo Group Title 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

1. tukajo Group Title

@experimentX

2. tukajo Group Title

@Chlorophyll

3. experimentX Group Title

0 to 0 this should be 0

4. tukajo Group Title

Nope lol

5. tukajo Group Title

[0,4]x[0,2] means a square from x = 0 to 4 and y = 0 to 2

6. tukajo Group Title

rectangle not square

7. tukajo Group Title

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 Group Title

|dw:1353065276769:dw|

9. tukajo Group Title

bottom one.

10. tukajo Group Title

Now how do I write those as limits of riemann sums?

11. experimentX Group Title

do the usual way ... first with respect to dx |dw:1353065464534:dw|

12. tukajo Group Title

book says this:|dw:1353065522208:dw|

13. tukajo Group Title

sorry for the pelletty drawing lol

14. experimentX Group Title

yeah you can do that!!

15. tukajo Group Title

But how do I write it out?

16. tukajo Group Title

Like the first step.. I guess..

17. experimentX Group Title

$x_i = {i \over N}, y_i = {i \over M}$ where N and M are your horizontal and vertical steps.

18. experimentX Group Title

$\Delta x = {1 \over N} \text{ and } \Delta y = {1\over M}$

19. tukajo Group Title

$\lim n ->\infty of \sum_{i = 1}^{n} \sum_{j = 1}^{n} f(x _{i^*j},y_{ij^*})\Delta A$

20. tukajo Group Title

Says use M = N for simplicity, where do I go from there?

21. experimentX Group Title

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 Group Title

but you are taking a limit as n approaches infinity for an exact answer...

23. tukajo Group Title

and what do I put in for f(xij,yij)?

24. experimentX Group Title

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 Group Title

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 Group Title

$\sum_{i=0}^{N}\sum_{j=0}^{N} (3(x_i)^2 - 2y_i - 2)\; \Delta x \Delta y$

27. experimentX Group Title

hold on!!

28. tukajo Group Title

haha sorry

29. experimentX Group Title

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 Group Title

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 Group Title

$\huge \Delta x = \frac{ 4 }{ N }, \Delta y = \frac{2}{N}$

32. experimentX Group Title

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 Group Title

$\lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 = \int_0^2 \int_0^4 (3x^2 - y - 2) dx dy$ I am not sure about your region.

34. tukajo Group Title

j=1 and i = 1, not 0... so (3(1/n)^2 -2(1/n) -2)*(1/n)^2

35. experimentX Group Title

must be something similar ... just fix your integration.

36. tukajo Group Title

well I mean if you plugged in 0 for i and j wouldn't it erase the values?

37. experimentX Group Title

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 Group Title
39. tukajo Group Title

I think I'll ask my professor tomorrow. Sorry, I'm a bit lost. I appreciate the help.

40. experimentX Group Title

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 Group Title

what a mess

42. Zarkon Group Title

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