anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Graph it. Looks like it is bounded from x=0 to x=2 \[\int\limits_{0}^{2}\sin^{-1} (x/2)\]
anonymous
  • anonymous
yes 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
  • anonymous
respect to y means leaving the y alone You are referring to respect to x

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anonymous
  • anonymous
um but it sort of says integrate with respect to y.... that's the part that im confused about
anonymous
  • anonymous
read what I wrote above
anonymous
  • anonymous
lol yea i read it.
anonymous
  • anonymous
I 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
  • anonymous
If 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
  • anonymous
It 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
  • anonymous
Since it respect to y, the f(x) does not have to change..
anonymous
  • anonymous
the y=f(x) function
anonymous
  • anonymous
okay so then would it be \[\int\limits\limits_{-1}^{0}\ \frac{ \sin(y) }{ 2 } dy\]
anonymous
  • anonymous
am i completely wrong ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Graph 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.
thomas5267
  • thomas5267
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anonymous
  • anonymous
the boundaries|dw:1438909739801:dw| are 0 and 2 right
anonymous
  • anonymous
well @thomas5267 picture is better but you get the point.
anonymous
  • anonymous
Its actually \[\int\limits_{0}^{1.5}whatever the inverse function of y=\arcsin(x/2) is\]
anonymous
  • anonymous
why 1.5 ?
anonymous
  • anonymous
for 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
  • anonymous
approaching 2 y=1.5
anonymous
  • anonymous
coordinates are around (2,1.5)
anonymous
  • anonymous
oh i see, okay that makes sense how do i find the inverse
anonymous
  • anonymous
y=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
  • anonymous
tell me if this is wrong\[\int\limits\limits\limits_{0}^{1.5}\ \frac{ \sin(y) }{ 2 } dy\]
anonymous
  • anonymous
It 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
  • anonymous
okay no worris i can probably find the inverse with some reasearch. Thank's for the help!
anonymous
  • anonymous
just remember that the upper and lower limit of integration changes when the x=f(y) or y=f(x) changes. No problem!
anonymous
  • anonymous
. I made a mistake. You have to get x by itself [x=f(y)]. This is not called an inverse.

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