anonymous
  • anonymous
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
where are the curves?
anonymous
  • anonymous
im attaching a file :) it had less than equal to signs and what not, so i just thought id attach the file
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anonymous
  • anonymous
Okay so we first plot the functions

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anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=Plot%5B{5-4cos%5Bx%5D%2Ccos%5Bx%5D}%5D{x%2C0%2CPi%2F2}%5D
anonymous
  • anonymous
whoa, my graph was way off
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anonymous
  • anonymous
no I think your right too
anonymous
  • anonymous
ohh i see your graph has a bigger scale xD
anonymous
  • anonymous
Yes, exactly
anonymous
  • anonymous
okay, so can we set the two equations equal to eachother, to find the boundaries?
anonymous
  • anonymous
actually, if we observe the graph, you see that they don't seem to have any intersection point between the interval 0, pi/2
anonymous
  • anonymous
ooh i thought it intersected near the very beginning cuz it was so close lol
anonymous
  • anonymous
Hold on
anonymous
  • anonymous
kay. normally to find the boundaries of the two curves, you'd set them equal to eachother right? thats what i did with this problem and i ended up getting x = 0, and 2pi
anonymous
  • anonymous
you are right it intersect at 0 http://www.wolframalpha.com/input/?i=Solve%5B5-4cos%5Bx%5D%3D%3Dcos%5Bx%5D%2Cx%5D
anonymous
  • anonymous
The second part ask, should you integrate with respect to x or y what do you think?
anonymous
  • anonymous
i think with respect to x. i honestly dont really know the difference, i usually integrate with respect to x lol ><
anonymous
  • anonymous
Okay,it is a matter of convenience, you can do with either
anonymous
  • anonymous
If we were to integrate with respect to x we can see that cos[x] is always below the other function
anonymous
  • anonymous
However,it get tricker when you try to integrate with respect to y because for all y<1 you will need one integral and all y>1 you will need another integral Do you see why?
anonymous
  • anonymous
is it because the y values for the upper (positive values) differ from the bottom (negative values) ?
anonymous
  • anonymous
oh wait. is it because the highest y value cosx can go is either -1 or 1
anonymous
  • anonymous
and for the other function, it can go higher than that
anonymous
  • anonymous
When you are integrating with respect to x what you are doing is adding up infinite number of verticle strip between two functions from x=0 to x=pi/2
anonymous
  • anonymous
how about when you integrate with respect to y?
anonymous
  • anonymous
Go here http://www.twiddla.com/537036

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