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anonymous
 5 years ago
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.
anonymous
 5 years ago
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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where are the curves?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im attaching a file :) it had less than equal to signs and what not, so i just thought id attach the file

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay so we first plot the functions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=Plot%5B {54cos%5Bx%5D%2Ccos%5Bx%5D}%5D{x%2C0%2CPi%2F2}%5D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoa, my graph was way off

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no I think your right too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh i see your graph has a bigger scale xD

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, so can we set the two equations equal to eachother, to find the boundaries?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, if we observe the graph, you see that they don't seem to have any intersection point between the interval 0, pi/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ooh i thought it intersected near the very beginning cuz it was so close lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kay. 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
 5 years ago
Best ResponseYou've already chosen the best response.0you are right it intersect at 0 http://www.wolframalpha.com/input/?i=Solve%5B54cos%5Bx%5D%3D%3Dcos%5Bx%5D%2Cx%5D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The second part ask, should you integrate with respect to x or y what do you think?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think with respect to x. i honestly dont really know the difference, i usually integrate with respect to x lol ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay,it is a matter of convenience, you can do with either

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If we were to integrate with respect to x we can see that cos[x] is always below the other function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0However,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
 5 years ago
Best ResponseYou've already chosen the best response.0is it because the y values for the upper (positive values) differ from the bottom (negative values) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait. is it because the highest y value cosx can go is either 1 or 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and for the other function, it can go higher than that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When 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
 5 years ago
Best ResponseYou've already chosen the best response.0how about when you integrate with respect to y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Go here http://www.twiddla.com/537036
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