Asked by Nick
Sketch the region in the first quadrant enclosed by y=6/x, y=2x, and y=.5x. Decide whether to integrate with respect to x or y. Then find the area of the region.
need final area as answer please, and step would be helpful also if you can post.
need final area as answer please, and step would be helpful also if you can post.
Answers
Answered by
Damon
sketch it of course
where does y = 2x hit y=6/x?
2 x = 6/x
x^2 = 3
x = sqrt 3
then y = 2 sqrt 3
so
from x = 0 to x = sqrt 3 integrate y = (2 x-.5 x) = 1.5 x
1.5 x^2/2
.75 ( 3-0) = 2.25
then where does y = .5 x hit y = 6/x?
.5 = 6/x^2
x^2 = 12
x = 2 sqrt 3
so integrate y = (6/x - .5 x) from x = sqt 3 to x = 2 sqrt 3
(6 ln x - x^2/4)
6 ln (2 sqrt 3 - sqrt 3) - .25(12-3)
2.62
so 2.25 + 2.62
where does y = 2x hit y=6/x?
2 x = 6/x
x^2 = 3
x = sqrt 3
then y = 2 sqrt 3
so
from x = 0 to x = sqrt 3 integrate y = (2 x-.5 x) = 1.5 x
1.5 x^2/2
.75 ( 3-0) = 2.25
then where does y = .5 x hit y = 6/x?
.5 = 6/x^2
x^2 = 12
x = 2 sqrt 3
so integrate y = (6/x - .5 x) from x = sqt 3 to x = 2 sqrt 3
(6 ln x - x^2/4)
6 ln (2 sqrt 3 - sqrt 3) - .25(12-3)
2.62
so 2.25 + 2.62
Answered by
Reiny
Nick, with a little effort on your part to complete the solution I gave you and some small adjustment for this question, you could have saved
the duplication of this solution by Damon.
the duplication of this solution by Damon.
Answered by
Damon
Oh my :(
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