Line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions.

Question

There is a line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions. Find m.

My solution:

When I compute my answer, I get $1-\frac{1}{\sqrt[3]{2}}$ (as seen in Wolfram below)

Solution in Wolfram:

https://www.wolframalpha.com/input/?i=%281-x%29%5E2%28-2x%2B2%29%3D1

Solution in Geogebra:

https://www.geogebra.org/calculator/kshvr8mw

My question: is there a way to find this solution without a GDC/Geogebra? In other words, a method which would lead to a simpler equation in the end?

Thank you.

Answer 1

This is my solution.

First, we recognize that the quadratic has roots at $(0,0)$ and $(1,0)$. So, to first solve for the area under the quadratic:

$\int_0^1 x-x^2.dx=\left[\frac{x}{2}-\frac{x^3}{3}\right]_0^1$

$=\frac{1}{6}$

So, half of this area is $\frac{1}{12}$

Let $x=a$ be the abscissa of the intersection point of line $y=mx$ and curve $y=x(1-x)$ different from $0$.

Then:

$ma=a(1-a)$

$m=(1-a)$

So, the equation of the straight line is $y=(1-a)x$

$\int_0^a(x-x^2)-(1-a)x.dx=\frac{1}{12}$

$\int_0^a-x^2+ax.dx=\frac{1}{12}$

$\left[-\frac{x^3}{3}+\frac{ax^2}{2}\right]_0^a=\frac{1}{12}$

$-\frac{a^3}{3}+\frac{a^3}{2}=\frac{1}{12}$

$\frac{a^3}{6}=\frac{1}{12}$

$a^3=\frac{1}{2}$

$a=\sqrt[3]{\frac{1}{2}}$

As earlier determined, the equation of the straight line is $y=(1-a)x$, so $y=(1-\sqrt[3]{\frac{1}{2}})x$.

Answer 2

You can simplify further from your last step,

$$(1-m)^2(-2m+2)=1$$ $$(1-m)^22(1-m)=1$$ $$(1-m)^3 = \frac{1}{2}$$ $$1-m = \frac{1}{\sqrt[3]{2}}$$ $$m=1-\frac{1}{\sqrt[3]{2}}$$