Asked by McKenna
a) Find the area of the region R bounded by the graphs of the equations y=2x−x^2, x=0, and y=0.
b)The line y=cx, where c>0, divides R into two subregions. Compute the value of c for which the two subregions have the same area.
b)The line y=cx, where c>0, divides R into two subregions. Compute the value of c for which the two subregions have the same area.
Answers
Answered by
Steve
the x=0 boundary is redundant, since the parabola intersects the x-axis at 0 and 2.
Anyway, the area is just a bunch of vertical strips of width dx and height y, so
a = ∫y dx
= ∫[0,2] 2x-x^2
= 4/3
Now, you want the line to divide R into two equal areas. y=cx intersects the parabola at
2x-x^2 = cx
x^2+(c-2)x = 0
x(x+c-2)=0
x = 0, x = 2-c
Since the boundary changes at 2-c, we need
∫[0,2-c] 2x-x^2-cx dx
= ∫[0,2-c] cx dx + ∫[2-c,2] 2x-x^2 dx
(2-c)^3/6 = c(2-c)^2/2 + c^2(3-c)/3
c^3-6c^2+12c+4 = 0
c = (2-∛4)
Anyway, the area is just a bunch of vertical strips of width dx and height y, so
a = ∫y dx
= ∫[0,2] 2x-x^2
= 4/3
Now, you want the line to divide R into two equal areas. y=cx intersects the parabola at
2x-x^2 = cx
x^2+(c-2)x = 0
x(x+c-2)=0
x = 0, x = 2-c
Since the boundary changes at 2-c, we need
∫[0,2-c] 2x-x^2-cx dx
= ∫[0,2-c] cx dx + ∫[2-c,2] 2x-x^2 dx
(2-c)^3/6 = c(2-c)^2/2 + c^2(3-c)/3
c^3-6c^2+12c+4 = 0
c = (2-∛4)
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