# 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
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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