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anonymous
 one year ago
Find the area of the region bounded by the curves y =arcsin(x/2), y = 0, and x = 2 obtained by integrating with respect to y.
anonymous
 one year ago
Find the area of the region bounded by the curves y =arcsin(x/2), y = 0, and x = 2 obtained by integrating with respect to y.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Graph it. Looks like it is bounded from x=0 to x=2 \[\int\limits_{0}^{2}\sin^{1} (x/2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but what do i do about integrating with respect to y, i thought that meant you can only have y expressions in the integrant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0respect to y means leaving the y alone You are referring to respect to x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0um but it sort of says integrate with respect to y.... that's the part that im confused about

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0read what I wrote above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know this will give me the area \int\limits_{0}^{2}\sin^{1} (x/2) dx im just not sure how to find that same area but with respect to y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If it is respect to y then you do not change the function. y=arcsin(x/2) does not need to be change do the x= form The integration boundaries are the range of the x coordinates

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It means the same thing. If it is respect to x. It would have about 2 x=f(y) functions but you have to find the inverse of it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since it respect to y, the f(x) does not have to change..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so then would it be \[\int\limits\limits_{1}^{0}\ \frac{ \sin(y) }{ 2 } dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0am i completely wrong ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Graph the function and tell me the area bounded at the x axis. That is respect to y Respect to x is finding the area bounded at the y axis and converting y=arcsin(x/2) into x= ? which is the inverse function.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the boundariesdw:1438909739801:dw are 0 and 2 right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well @thomas5267 picture is better but you get the point.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its actually \[\int\limits_{0}^{1.5}whatever the inverse function of y=\arcsin(x/2) is\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for respect to y you care about x axis so its bounded from x= 0 to x= 2 for respect to x you care about y axis so its bounded from y=0 to y= around 1.5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0coordinates are around (2,1.5)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i see, okay that makes sense how do i find the inverse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y=arcsin(x/2) sin(y)=x/2 x=sin(y)/2 I think. I'm not sure. You are not supposed to find the inverse to this problem anyways, since it is a bit challenging. Unless you can use a calculator.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tell me if this is wrong\[\int\limits\limits\limits_{0}^{1.5}\ \frac{ \sin(y) }{ 2 } dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is correct according to me. But i do not know if the find the inverse function of arcsin(x/2) correctly, but I am pretty sure.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay no worris i can probably find the inverse with some reasearch. Thank's for the help!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just remember that the upper and lower limit of integration changes when the x=f(y) or y=f(x) changes. No problem!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0. I made a mistake. You have to get x by itself [x=f(y)]. This is not called an inverse.
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