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.

Question

need final area as answer please, and step would be helpful also if you can post.

Answer 1

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

Answer 2

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.

Answer 3

Oh my :(