Ac3
  • Ac3
Find the average value of the function f over the region R. f(x,y)=4x+10y. R is the region bounded by the coordinate axes and the lines x+y=2 and x+y=5.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Ac3
  • Ac3
@ganeshie8
Ac3
  • Ac3
@IrishBoy123 think you can help me set up this integral?
IrishBoy123
  • IrishBoy123
sure what you got so far?

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Ac3
  • Ac3
This is a problem from a test. And I got it wrong. :/
Ac3
  • Ac3
This is what I thought it was.
Ac3
  • Ac3
|dw:1449766838694:dw|
Ac3
  • Ac3
and so when we're trying to find the "average value" we do....
Ac3
  • Ac3
\[\frac{ 1 }{ area of the Region }\int\limits_{a}^{b}\int\limits_{c}^{d}f(x,y)dydx\]
Ac3
  • Ac3
the order of integration doesn't necessarily need to by dydx
IrishBoy123
  • IrishBoy123
yes, that should be the broad approach. how's about the integral?? i would be tempted to do \(A \bar f = \int\limits_{y=0}^{5} \int\limits_{x=0}^{5-y} \; dx \; dy \; (4x +10y) - \int\limits_{y=0}^{2} \int\limits_{x=0}^{2-y} \; dx \; dy \; (4x +10y)\) where A is the area of that region might just see if i can stuff that into wolfram....
Ac3
  • Ac3
My integral was.... \[\frac{ 1 }{ 5 }\int\limits_{0}^{3}\int\limits_{0}^{2}4x+10ydydx\]
IrishBoy123
  • IrishBoy123
http://www.wolframalpha.com/input/?i=int_%7By%3D0%7D%5E%7B5%7D+int%5Climits_%7Bx%3D0%7D%5E%7B5-y%7D+%284x+%2B10y%29+dx+++dy+++-++int_%7By%3D0%7D%5E%7B2%7D+int_%7Bx%3D0%7D%5E%7B2-y%7D+%284x+%2B10y%29+dx++dy++
IrishBoy123
  • IrishBoy123
so divide that by 21/2
Ac3
  • Ac3
The integral my teacher told me it was, was.... \[\frac{ 1 }{ \frac{ 21 }{ 2 } }\int\limits_{0}^{2}\int\limits_{2-x}^{5-x}4x+10ydydx+\int\limits_{2}^{5}\int\limits_{0}^{5-x}4x+10ydydx\]
IrishBoy123
  • IrishBoy123
ah! http://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx
Ac3
  • Ac3
that 21/2 I can't see how to get it.
Ac3
  • Ac3
I could've swore the area was 5
IrishBoy123
  • IrishBoy123
yes, your teacher and mine are same, except i did dx dy, teacher did dydx symmetry so you can see the pattern \(\frac{1}{2}5*5 - \frac{1}{2}2*2\) the area of the region
IrishBoy123
  • IrishBoy123
\[\frac{1}{2}5*5 - \frac{1}{2}2*2\color{red}{=21/2}\]
IrishBoy123
  • IrishBoy123
your integral, ie \(\frac{ 1 }{ 5 }\int\limits_{0}^{3}\int\limits_{0}^{2}4x+10ydydx\) is for this region |dw:1449767608693:dw|
Ac3
  • Ac3
lol
IrishBoy123
  • IrishBoy123
that Pauls Online Math holds the key to getting the region right in the integration limits and you have to do the bigger triangle less the smaller triangle to get the right region
Ac3
  • Ac3
wait the triangles aren't equal?
Ac3
  • Ac3
when I was finding the area this is how I decided to go about it.
Ac3
  • Ac3
|dw:1449767796916:dw|
IrishBoy123
  • IrishBoy123
should be this way |dw:1449767952913:dw| MINUS |dw:1449767973736:dw| 1/2 * 5 * 5 - 1/2 * 2 * 2 = 21/2
Ac3
  • Ac3
dammm i didn't think we needed to include that.
Ac3
  • Ac3
I could've swore it was just this region..
Ac3
  • Ac3
|dw:1449768076660:dw|
Ac3
  • Ac3
Ok so I found my first mistake. And you said that website will show me how to get the bound of integration?
IrishBoy123
  • IrishBoy123
Paul's and this video is good once you know what you are looking for https://www.youtube.com/watch?v=p7xhvJGyjKE
Ac3
  • Ac3
thx man
Ac3
  • Ac3
I got one more question should be a lot quicker though.
IrishBoy123
  • IrishBoy123
pleasure best open new thread, i may not be able to help,,,,,, if it is hard :-)

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