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. πŸ‘ 0
  2. πŸ‘Ž 0
  3. πŸ‘ 133
asked by Henry

  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 (√(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

    1. πŸ‘ 0
    2. πŸ‘Ž 0
    posted by Steve

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