# Find the x−coordinate of the point A

• Oct 7th 2010, 02:47 AM
blackdragon190
Find the x−coordinate of the point A
I'm given the function $\displaystyle f(x)=-3x(x-2)$ and must work out the x coordinate on f(x), A, which a line dividing the area between the function and the x-axis from [0,2] into 2 parts of equal area passes through. The area total area between the function and x-axis from [0,2] is 4 square units.

I've got it down to $\displaystyle -A^3+3A^2-2=0$

I'm kinda stuck there. Is there another way to do it that doesn't require me to solve polynomial equations?
• Oct 7th 2010, 03:05 AM
mr fantastic
Quote:

Originally Posted by blackdragon190
I'm given the function $\displaystyle f(x)=-3x(x-2)$ and must work out the x coordinate on f(x), A, which a line dividing the area between the function and the x-axis from [0,2] into 2 parts of equal area passes through. The area total area between the function and x-axis from [0,2] is 4 square units.

I've got it down to $\displaystyle -A^3+3A^2-2=0$

I'm kinda stuck there. Is there another way to do it that doesn't require me to solve polynomial equations?

Assuming your calculations are correct, it's not difficult to see that A = 1 is a possible solution ....
• Oct 7th 2010, 03:06 AM
Quote:

Originally Posted by blackdragon190
I'm given the function $\displaystyle f(x)=-3x(x-2)$ and must work out the x coordinate on f(x), A, which a line dividing the area between the function and the x-axis from [0,2] into 2 parts of equal area passes through. The area total area between the function and x-axis from [0,2] is 4 square units.

I've got it down to $\displaystyle -A^3+3A^2-2=0$

I'm kinda stuck there. Is there another way to do it that doesn't require me to solve polynomial equations?

Does the line have to pass through a certain point on the x-axis?
Numerous lines will divide the area in two equal parts.
The most obvious one is the line x=1, since the graph is symmetrical about x=1.
• Oct 7th 2010, 03:08 AM
Also sprach Zarathustra
$\displaystyle -a^3+3a^2-2=0=-(a-1) (a^2-2 a-2)$
• Oct 7th 2010, 03:25 AM
blackdragon190
I don't think I expressed the problem well enough...here's a picture of what I mean:

http://img.photobucket.com/albums/v6...g002/maths.jpg

Argh, I just realized that my initial calculation is totally wrong.
• Oct 7th 2010, 04:16 AM
HallsofIvy
Let $\displaystyle (x_0,-3(x_0-1)^2+ 3)$ be the point where that line crosses the parabola. Then the line has equation $\displaystyle y= \frac{-3(x_0-1)^2+ 3}{x_0}x$ and the area, above that line but below the parabola is given by
$\displaystyle \int_0^x_0 -3x^2+ 6+ 3\frac{(x_0-1)^2- 3}{x_0}x dx$
Since the area under the entire parabola is 4, find that integral, in terms of $\displaystyle x_0$, set it equal to 2, and solve for $\displaystyle x_0$.
• Oct 7th 2010, 05:58 AM
Quote:

Originally Posted by blackdragon190
I don't think I expressed the problem well enough...here's a picture of what I mean:

http://img.photobucket.com/albums/v6...g002/maths.jpg

Argh, I just realized that my initial calculation is totally wrong.

If you integrate to get the area under the parabola from x=0 to x=A,
then subtract the area of the triangle, that resulting area is 2.

The area of the triangle is $\displaystyle \displaystyle\frac{Af(A)}{2}$

Alternatively, integrate from x=A to x=2 and add the triangle area.
Those combined areas are also 2.

$\displaystyle f(x)=-3x^2+6x$

$\displaystyle f(A)=-3A^2+6A$

$\displaystyle \displaystyle\int_{0}^Af(x)}dx=2+\frac{Af(A)}{2}$

$\displaystyle \displaystyle\int_{0}^A\left(-3x^2+6x\right)}dx=2+\frac{-3A^3+6A^2}{2}$

$\displaystyle A^3=4$