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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.

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  1. anonymous
    • one year ago
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    Graph it. Looks like it is bounded from x=0 to x=2 \[\int\limits_{0}^{2}\sin^{-1} (x/2)\]

  2. anonymous
    • one year ago
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    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

  3. anonymous
    • one year ago
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    respect to y means leaving the y alone You are referring to respect to x

  4. anonymous
    • one year ago
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    um but it sort of says integrate with respect to y.... that's the part that im confused about

  5. anonymous
    • one year ago
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    read what I wrote above

  6. anonymous
    • one year ago
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    lol yea i read it.

  7. anonymous
    • one year ago
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    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

  8. anonymous
    • one year ago
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    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

  9. anonymous
    • one year ago
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    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.

  10. anonymous
    • one year ago
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    Since it respect to y, the f(x) does not have to change..

  11. anonymous
    • one year ago
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    the y=f(x) function

  12. anonymous
    • one year ago
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    okay so then would it be \[\int\limits\limits_{-1}^{0}\ \frac{ \sin(y) }{ 2 } dy\]

  13. anonymous
    • one year ago
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    am i completely wrong ?

  14. anonymous
    • one year ago
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    Yes

  15. anonymous
    • one year ago
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    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.

  16. thomas5267
    • one year ago
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  17. anonymous
    • one year ago
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    the boundaries|dw:1438909739801:dw| are 0 and 2 right

  18. anonymous
    • one year ago
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    well @thomas5267 picture is better but you get the point.

  19. anonymous
    • one year ago
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    Its actually \[\int\limits_{0}^{1.5}whatever the inverse function of y=\arcsin(x/2) is\]

  20. anonymous
    • one year ago
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    why 1.5 ?

  21. anonymous
    • one year ago
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    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

  22. anonymous
    • one year ago
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    approaching 2 y=1.5

  23. anonymous
    • one year ago
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    coordinates are around (2,1.5)

  24. anonymous
    • one year ago
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    oh i see, okay that makes sense how do i find the inverse

  25. anonymous
    • one year ago
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    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.

  26. anonymous
    • one year ago
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    tell me if this is wrong\[\int\limits\limits\limits_{0}^{1.5}\ \frac{ \sin(y) }{ 2 } dy\]

  27. anonymous
    • one year ago
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    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.

  28. anonymous
    • one year ago
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    okay no worris i can probably find the inverse with some reasearch. Thank's for the help!

  29. anonymous
    • one year ago
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    just remember that the upper and lower limit of integration changes when the x=f(y) or y=f(x) changes. No problem!

  30. anonymous
    • one year ago
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    . I made a mistake. You have to get x by itself [x=f(y)]. This is not called an inverse.

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