math, integral and analytical problem?

Hello

The parabola has its vertex at (3, 9), and has roots at x = 0 and 6.

You have to find m of the line y = mx dividing the area into equal halves.

Now find the intersection point of the straight line with the parabola:

- x^2 + 6x = mx

x^2 - x(6 - m) = 0

x = 6 - m

To define the upper half of the parabola is easy:

A(up) = ∫(-x^2 + 6x - mx)dx from x = 0 to x = (6 - m).

The lower half of the parabola is the integral of the line from x = 0 to 6 minus the area under the line outside of the parabola.

This area outside of the parabola is ∫(mx - (- x^2 + 6x))dx from x = (6 - m) to x = 6.

And the area of the parabola under the line is then

A(lo) = ∫(mxdx from x = 0 to 6 minus ∫(mx - (- x^2 + 6x))dx from x = (6 - m) to x = 6.

Now set the two areas equal:

∫(-x^2 + 6x - mx)dx (from x = 0 to x = (6-m)) = ∫(mxdx (from x = 0 to 6) minus ∫(mx + x^2 - 6x))dx (from x = (6 - m) to x = 6).

[- 1/3*x^3 + 3x^2 - mx^2/2](from 0 to 6) = [mx^2/2] (from 0 to 6) - [mx^2/2 +x^3/3 - 3x^2] (from (6-m) to 6).

plug the borders in, and solve for m, which I do not want to show here.

It boils down to

5/6*m^3 - m^3/3 + 36m - 36 = 0 (if I had no error in my calculation)

The solution (Wolfram alpha) is

m = 0,986774

The equation of your line is

y = 0,986774 x

Regards

x = -b / 2a

x= -6/-2

x= 3

the line is x = 3