Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
Sorry, I translated it from the german "Parabel". So in german it is still called this way. –  TestGuest Apr 26 '13 at 7:07
1  
This picture may help. wolframalpha.com/input/… –  in_wolframAlpha_we_trust Apr 26 '13 at 7:25

1 Answer 1

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).

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.