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.
y=3+√X, y=3+1/5x What is the area?

  1. 0
asked by Nick
  1. Looks pretty straightforward, both equations say y = ...

    let's find their intersection:

    3 + √x = 3 + (1/5)x
    √x = x/5
    square both sides,
    x = x^2/25
    25x - x^2 = 0
    x(25-x) = 0
    x = 0 or x = 25

    if x = 0, y = 3
    if x = 25, y = 8

    effective height of regions
    = 3+√x - 3 - (1/5)x
    = x^(1/2) - (1/5)x

    area = ∫(x^(1/2) - (1/5)x) dx from x = 0 to 25
    = [ (2/3)x^(3/2) - (1/10)x^2 ] from 0 to 25
    = ( (2/3)(125) - (1/10)(625) - 0)
    = 125/6
    = appr 20.833

    posted by Reiny

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