## anonymous 3 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)

1. anonymous

@experimentX

2. anonymous

@Chlorophyll

3. experimentX

0 to 0 this should be 0

4. anonymous

Nope lol

5. anonymous

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

6. anonymous

rectangle not square

7. 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.

8. experimentX

|dw:1353065276769:dw|

9. anonymous

bottom one.

10. anonymous

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

11. experimentX

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

12. anonymous

book says this:|dw:1353065522208:dw|

13. anonymous

sorry for the pelletty drawing lol

14. experimentX

yeah you can do that!!

15. anonymous

But how do I write it out?

16. anonymous

Like the first step.. I guess..

17. experimentX

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

18. experimentX

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

19. anonymous

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

20. anonymous

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

21. 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.

22. anonymous

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

23. anonymous

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

24. experimentX

this is numerical integration, just replace x and y by x_i and y_i in f(x,y) $x_i = {i \over N}, y_i = {i \over M}$

25. 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?

26. experimentX

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

27. experimentX

hold on!!

28. anonymous

haha sorry

29. 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

30. 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}]

31. anonymous

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

32. experimentX

write it as $\lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2$

33. 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.

34. anonymous

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

35. experimentX

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

36. anonymous

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

37. 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

38. experimentX
39. anonymous

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

40. 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.

41. Zarkon

what a mess

42. 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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