integration of sin . (1 Viewer)

[apak]

find the area btwn y=sinx and y=1/2
for this question what are the limits?
0-30, 30-150, 150 to 360
or 0-30, 30-150, 150-180, 180 to 360?
also do you have to absolute the answer? if so how do you absolute things such as -(5/12)pi + root3/2
and also how do you solve the question
thx in adv

 

[sasquatch]

In your question you stated what is the area bound by y=sinx and y=(1/2), so that only includes areas which are completly between the two lines.

Im assuming the range is 0 <= x <= 2π, but if it wasnt, there would be more area between both curves then what follows.

Look at your first diagram, that is the correct interval, which you can figure out as the following

[1] y = sin x
[2] y = (1/2)

Sub [1] into [2]
sinx = (1/2)
x = π/6 or 5π/6

*NOTE*: With intergration and differentiation you must ALWAYS work in radians, using degrees is wrong.

Your question regarding "do you need to absoulte anything". You only need to absolute the entire definate integral when you are unsure which curve lies above the other. In this case you can clearly see that y = sin x, lies above y = (1/2) so it is neccessary to subtract the area under the cuve y=(1/2) in the interval specified above, fromthe area under the curve y = sin x in the same interval. This will ensure your answer is positive.

This will look ugly, if you cant follow i could write it up in a word document, but for now,

A = _π/6∫^5π/6 [sinx-(1/2)]dx
= [ -cos x - (1/2)x ]_π/6^5π/6
= [- cos(5π/6) - (5π/12)] - [-cos(π/6)-(π/12]
= (√3/2) - (5π/12) - [-(√3/2)-(π/12]
= (√3/2) - (5π/12) + (√3/2) + (π/12
= √3 - (4π/12)
= [√3 - (π/3)] units²