Intergration Problem: eqn of line dividing region equally

[reehlgirl1986]

The problem states:

Let "R" be the region bounded by the parabola y = x - x² and the x-axis. Find the equation of the line y = mx that divides the region into two regions of equal area.

1. I found the area of x - x² to the x-axis to be 1/6 when I intergrated from zero to one, so the area of x - x² to mx has to equal 1/12.

2. I found the where mx = x - x², at 0 and when x = -(m + 1).

3. I intergrateed from 0 to some number ((x - x²) - (mx) which equals (1/2x²) - (1/3x³) - (1/2mx²)

4. I put "-m + 1" for "x".

. . .(1/2m + 1²) - (-1/3m + 1³) - (1/2m - m + 1²) = 1/12

5. This equals:

. . .(-1/2m²) + (1/2) + (1/3m²) - (1/3) + (1/2m³) + (1/2m) = 1/12

6. I sub the 1/2 and + 1/3 to the 1/12 to equal -1/12

7. I took:

. . .(-1/2m²) + (1/3m²) + (1/2m³) + (1/2m) = -1/12

...and divided the whole thing by negative 1/2 to equal:

. . .(m²) + (2/3m²) - (m³) - (m) = 1/6

QUESTION: Can I combine these or not not? For instance, (m³) and (m) to equal m², and then add m² twice to equal 2m², and then add 2/3m² to equal 8/3m²? If so, I can then divide 1/6 by 8/3 and square root it to get "0.25" as "m". But when I do that, I get "0.07" as the area, which is close but not 0.083333333.

Where did i go wrong? Please help!

 

[royhaas]

The quadratic intersects the line at \(\displaystyle x = 1-m\).

 

[reehlgirl1986]

It still does not equal 1/12 when you work it with m-1, but thank you for trying.

 

[pka]

As Prof Haas has pointed out the intersection point is (1-m,[1-m]-[1-m]<SUP>2</SUP>].

So solve the equation for m: \(\displaystyle \L
\int\limits_0^{1 - m} {\left[ {\left( {x - x^2 } \right) - \left( {mx} \right)} \right]} dx = \frac{1}{2}\int\limits_0^1 {\left[ {\left( {x - x^2 } \right)} \right]} dx\)

 

[reehlgirl1986]

Nope!

When you put m-1 or -1+m in for x-you get answer, but not the right one the answer does not make the graphs halfed. Please more help!

 

[pka]

As you can see m=0.206 works.