Calculus

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y = 4(x^(1/2)), y=4, and 2y +2x = 6

I keep getting an area around 21.3 but it is incorrect. Am I close?

Thank you!

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asked by Donna
  1. odd that one function is written as

    2y = 4√x

    why not just y = 2√x ?

    Anyway, assuming the functions are correctly written, the area is roughly triangular, with vertices at (-1,4),(1,2),(4,4)

    Integrating along x, we need to divide it up into two regions, over [-1,1] and [1,4]

    a = ∫[-1,1] 4 - (3-x) dx + ∫[1,4] 4 - 2√x dx
    = (x + 1/2 x^2)[-1,1] + (4x - 4/3 x^(3/2))[1,4]
    = 2 + 8/3
    = 14/3

    Integrating along y is easier, since we just subtract the left from the right:

    a = ∫[2,4] y^2/4 - (3-y) dy
    = y^3/12 + y^2/2 - 3y [2,4]
    = 14/3

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    posted by Steve

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