Integration with two unknowns I'm completely stumped with this one, I'm not sure how I should do this.
The equation of a parabola is $y=-3x(x-2)$. It intersects the $x$-axis at $0$ and $2$.
Given that the area of this parabola is $4\,{\rm units}^2$, there will be a straight line $y=mx$ which divides the area exactly in half ($2\,{\rm units}^2$ per half).
I need to find the $x$-coordinate (point $T$) of where the straight line and the parabola intersect (point $G$) - the $x$-coordinate of the point which divides the parabola into equal areas.
So far I've worked out that the gradient of the dividing line is $m = 6-3p$
I think what I have to do now is integrate a problem like this:
$$
\int\limits_0^T \big[ -3x(x-2)-(6-T)x \big] dx = 2
$$
(hope that formatted correctly)
Does anyone have any ideas?
Thanks,
John Smith
 A: A good idea would be to integrate $\max(f(x)-mx,0)$ between 0 and 2, which indeed leads you to solve $f(x)=mx$, which then gives you $x_M=2-\frac{m}{3}$. 
Now you simply have to integrate the following: $\int_0^{2-\frac{m}{3}} -3x(x-2) - mx dx = -\frac{1}{54}(m-6)^3$.
Making it equal to 2 gives you the result you're after.
A: Draw a picture. The intended region is the part of the parabola that is above the $x$axis. It indeed has area $4$.
Now let us find $m$. The line $y=mx$ meets the parabola where $mx=-3x^2+6x$. Beside the uninteresting solution $x=0$, we have the solution $x=\frac{6-m}{3}$. The corresponding $y$ is $\frac{m(6-m)}{3}$. Once we have found $m$, we will therefore be finished.  
I really don't much like fractions, so let $m=6q$. Then the line meets the parabola at $x=2-2q$.
Now we compute the area of the part of the parabolic region above $y=mx$. This is
$$\int_0^{2-2q} (-3x^2+6x-6qx)\,dx.$$
An antiderivative is $-x^3+3x^2-3qx^2$.  Substitute $x=2-2q$. We get 
$$(2-2q)^2[3-3q-(2-2q)]$$
which is $4(1-q)^3$. Set this equal to $2$ and solve for $q$.
