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Question

There is a line through the origin that divides the region bounded by the parabola y = 4 x - 2 x^2 and the x-axis into two regions with equal area. What is the slope of that line?

Answer 1

We have y = 4x - 2x^2 = 2x(2 - x)

This is 0 when either 2x = 0 or 2 - x = 0,

i.e. when x=0 or x=2.  This is a parabola

that opens downward since the coefficient

of x^2 is -2, a negative number.  That gives

you enough information to draw a good enough

picture of the curve.  The part above the

x-axis is between x=0 and x=2.

The equation of a line is y = mx+b, and since

this line goes through the origin, we have b=0,

so it's y = mx.  The other point at which the

line intersects the curve must be a point at

which mx = 2x(2 - x).  Since the other point

is at a place where x is not 0, we can divide

both sides by x.  We get m = 2(2 - x). Solving

that for x, we get x = 2 - (m/2).

The area below the curve and above the line

is the integral from 0 to 2 - (m/2), of the

height of the curve minus the height of the

line.  Thus it is the integral of 2x(2 - x) - mx

= 4x - 2x^2 - mx.  The antiderivative is

2x^2 - 2x^3/3 - mx^2/2.  If one plugs in

x = 0, one gets 0, and it remains to plug in

x = 2 - (m/2).  We get
2(2 - m/2)^2 - 2(2 - m/2)^3/3 -m(2 - m/2)^2/2.
With a bit of algebra this simplifies to

(1/3)(2 - m/2)^3.  That is the area above

the line and below the curve, and it must

be _half_ of the area between the cuve and

the x-axis.  So we need to know the area

between the curve and the x-axis.  That is

the integral from 0 to 2 of 4x - 2x^2.
That comes to 2x^2 - 2x^3/3 evaluated

at x = 2, or 8 - (16/3) = 8/3.  Half of that

is 4/3.

Therefore we need (1/3)(2 - m/2)^3 = 4/3.

Multiplying both sides by 3, we get

(2 - m/2)^3 = 4.

So 2 - m/2 = the cube root of 4.  Hence

m = 2(2 - cube root of 4).