# calculus

The region R is the region in the first quadrant bounded by the curves y = x^2-4x+4, x=0, and x=2.
Find a value h such that the vertical line x = h divided the region R into two regions of equal area. You get h = ?
I solved for the total area under the curve (8/3) and the area of each half (4/3). How do I get the h value from here?

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1. You are correct in the total area to be 8/3 and half of that would be 4/3
You also must have had the correct integral to obtain the 8/3
so just evaluate it from x = 0 to h instead of 0 to 2, then set that equal to 4/3

that is,
(1/3)(h^3) -2(h^2) + 4h - 0 = 4/3

This does not solve easily, unless I made an arithmetic error

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2. I think you forgot to include y=0 as part of the boundary,
and the line x=2 plays no role, since the vertex of the curve is at (2,0)
So, I'll go on assuming the triangular region bounded by the curve and the x- and y-axes.

As you say, ∫[0,2] (x^2-4x+4) dx = 8/3
So now you want
∫[0,h] (x^2-4x+4) dx = 4/3
That means that
1/3 h^3 - 2h^2 + 4h = 4/3
h^3-6h^2+12h-4 = 0
h^3-6h^2+12h-8 = -4
(h-2)^3 = -4
h-2 = -∛4
h = 2-∛4

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3. Thanks oobleck, my weakness is not reading questions
completely.

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