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
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.
3 answers
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.
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 ??
You changed one of your equations from
y = 12x to y = .5x
and you can't make the necessary changes in my solution ??