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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

martini
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user133422
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  • http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference

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     – Bman72 Jul 30 '14 at 15:24

2 Answers2

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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.

Matt B.
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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 $q$.

André Nicolas
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