Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
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....Asked by Henry
Sketch the region in the first quadrant enclosed by y=7/x, y=2x, and y=x/2. Decide whether to integrate with respect to x or y. Then find the area of the region.
Answers
Answered by
Steve
It doesn't really matter which variable is used, since the boundary changes in either direction.
Cleverly enough, the hyperbola intersects the two lines at (√(7/2),√14) and (√14,√(7/2).
So,
a = ∫[0,√(7/2)] (2x)-(x/2) dx
+ ∫[√(7/2),√14] (7/x)-(x/2) dx
or, along y,
a = ∫[0,√(7/2)] (2y)-(y/2) dy
+ ∫[0,√(7/2)] (7/y)-(y/2) dy
Cleverly enough, the hyperbola intersects the two lines at (√(7/2),√14) and (√14,√(7/2).
So,
a = ∫[0,√(7/2)] (2x)-(x/2) dx
+ ∫[√(7/2),√14] (7/x)-(x/2) dx
or, along y,
a = ∫[0,√(7/2)] (2y)-(y/2) dy
+ ∫[0,√(7/2)] (7/y)-(y/2) dy
There are no AI answers yet. The ability to request AI answers is coming soon!