Here's the question you clicked on:
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)
0 to 0 this should be 0
[0,4]x[0,2] means a square from x = 0 to 4 and y = 0 to 2
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.
|dw:1353065276769:dw|
Now how do I write those as limits of riemann sums?
do the usual way ... first with respect to dx |dw:1353065464534:dw|
book says this:|dw:1353065522208:dw|
sorry for the pelletty drawing lol
yeah you can do that!!
But how do I write it out?
Like the first step.. I guess..
\[x_i = {i \over N}, y_i = {i \over M}\] where N and M are your horizontal and vertical steps.
\[\Delta x = {1 \over N} \text{ and } \Delta y = {1\over M}\]
\[\lim n ->\infty of \sum_{i = 1}^{n} \sum_{j = 1}^{n} f(x _{i^*j},y_{ij^*})\Delta A\]
Says use M = N for simplicity, where do I go from there?
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.
but you are taking a limit as n approaches infinity for an exact answer...
and what do I put in for f(xij,yij)?
this is numerical integration, just replace x and y by x_i and y_i in f(x,y) \[ x_i = {i \over N}, y_i = {i \over M} \]
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?
\[ \sum_{i=0}^{N}\sum_{j=0}^{N} (3(x_i)^2 - 2y_i - 2)\; \Delta x \Delta y\]
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
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}]
\[\huge \Delta x = \frac{ 4 }{ N }, \Delta y = \frac{2}{N}\]
write it as \[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 \]
\[ \lim_{n \to \infty} \sum_{j=0}^{2n} \sum_{i=0}^{4n} \left (3 \left( i \over n\right )^2 - 2\left(j \over n\right) - 2\right)\left(1 \over n\right)^2 = \int_0^2 \int_0^4 (3x^2 - y - 2) dx dy\] I am not sure about your region.
j=1 and i = 1, not 0... so (3(1/n)^2 -2(1/n) -2)*(1/n)^2
must be something similar ... just fix your integration.
well I mean if you plugged in 0 for i and j wouldn't it erase the values?
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
I think I'll ask my professor tomorrow. Sorry, I'm a bit lost. I appreciate the help.
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.
\[\lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{m}f(x_{ij}^*,y_{ij}^*)\Delta A\] 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 \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[3\left(i\frac{4}{n}\right)^2-2\left(j\frac{2}{n}\right)-2\right]\frac{4}{n}\frac{2}{n}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[48\frac{i^2}{n^2}-4\frac{j}{n}-2\right]\frac{8}{n^2}\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\left[384\frac{i^2}{n^4}-32\frac{j}{n^3}-\frac{16}{n^2}\right]\] \[\lim_{n\to\infty}\sum_{i=1}^{n}\left[384\frac{i^2}{n^4}n-32\frac{1}{n^3}\frac{n(n+1)}{2}-\frac{16}{n^2}n\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^4}n\frac{n(n+1)(2n+1)}{6}-32\frac{1}{n^3}\frac{n(n+1)}{2}n-\frac{16}{n^2}nn\right]\] \[\lim_{n\to\infty}\left[384\frac{1}{n^2}\frac{(n+1)(2n+1)}{6}-32\frac{1}{n}\frac{(n+1)}{2}-16\right]\] \[=\frac{384\cdot 2}{6}-\frac{32}{2}-16=96\]