Calculus 2

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.

1. π 0
2. π 0
3. π 133
1. 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 (β(7/2),β14) and (β14,β(7/2).

So,

a = β«[0,β(7/2)] (2x)-(x/2) dx
+ β«[β(7/2),β14] (7/x)-(x/2) dx

or, along y,

a = β«[0,β(7/2)] (2y)-(y/2) dy
+ β«[0,β(7/2)] (7/y)-(y/2) dy

1. π 0
2. π 0
posted by Steve

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