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I need understanding a calculus question:

Johnny Bravo is building a head board with the outline of the $y = x - x^2$, which is bounded by the $x$-axis. He wants to split the head board into half green and half red, however, he does not want to split it straight down the middle. Johnny wants to draw a diagonal line starting at the origin that will divide the headboard in half. What is the equation of the line?

My approach was to try and find the intersection point (x-value) of the function that will set the two areas equal to each other.

I set up the integral equations:

$\int_{-1}^{x}x - x^2 dx = \int_{x}^{0}x-x^2 dx$

However, I do not think that this is the correct approach - it did not get me anywhere.

The correct answer is:

$y = (1 - \frac{1}{2^{1/3}})x$

Does anyone know how to arrive at this answer?

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    Draw a line starting at the origin? – 2017-02-23
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    The point (0, 0) must lie on the line. – 2017-02-23
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    You should try to make drawings. The way you did it, you'd be splitting the head board along a vertical line through the point $(x,0)$, rather than a sloped line through the origin $(0,0)$. Moreover, notice that the integration variable should not be the same variable that's showing up in the limits of integration. – 2017-02-23
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    your text says "original" just clarifying. – 2017-02-23
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    @DougM sorry, it's very late. I just corrected it. – 2017-02-23

1 Answers 1

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the area of the head board

$\int_0^1 x - x^2 \ dx = \frac 16$

line from the origin $y = kx$

intersects the curve $y = x - x^2$

$kx = x-x^2\\ (1-k) x - x^2 = 0$

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

$\frac {(1-k)^3}{6} = \frac 1 {12}$

$(1-k)^3 = \frac 12\\ k = 1-(\frac 12)^{\frac 13}\\ y = (1-(\frac 12)^{\frac 13})x$

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    This makes more sense - thanks! – 2017-02-23