Question

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.

Answers

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
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.
Damon
Oh my :(

Related Questions