# Area of intersection between parabola and square

The cubic $3x^2-x^3$ divides the square with endpoints A(0,0) and B(4,0) in three parts.

How can I show that this is true and what are the areas of the parts?

What I see so far, is that the parabola has roots at x = 0 and x = 3. But how can I prove that it cuts the square in three pieces (and how then to calculate the areas)?

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Sorry, I translated it from the german "Parabel". So in german it is still called this way. – TestGuest Apr 26 '13 at 7:07
This picture may help. wolframalpha.com/input/… – in_wolframAlpha_we_trust Apr 26 '13 at 7:25

Hints Plot the graph of the function, and draw the square. What changes when we consider $\dfrac1{10}(3x^2-x^3)$ instead?

For calculating the area, use suitably chosen integrals (this will require you to find the endpoints of the three parts).

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if we multiply it by 1/10, what changes is the range of the function on the y-scale (i.e. the maxima would be lower). How does that go into the direction of a proof? thanks – TestGuest Apr 26 '13 at 7:27
There will then only remain two parts... – Lord_Farin Apr 26 '13 at 7:28