Sketch the region in the first quadrant enclosed by y=6/x, y=2x, and y=12x. Decide whether to integrate with respect to x or y. Then find the area of the region.

Answer 1

intersection of y = 6/x with y = 2x
2x = 6/x
x^2 = 3
x = √3

intersection of y = 6/x with y = 12x
12x = 6/x
x^2 = 1/2
x = √2/2

According to my sketch I would find the area of the triangle from x = 0 to √2/2 with a height of
12x - 2x or

∫10x dx from 0 to √2/2

followed by the area of

∫(6/x - 2x) dx from x = √2/2 to √3

recall that integral 6/x is
6 lnx

Answer 2

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.

Answer 3

Good grief !

You changed one of your equations from
y = 12x to y = .5x

and you can't make the necessary changes in my solution ??