The Area of the region bounded by the x-axis and the graph of y= x^3-x: Is it:

1/2 ; 1/4 ; -1/4 ; 0

Question

The Area of the region bounded by the x-axis and the graph of y= x^3-x:  Is it:

1/2  ;  1/4  ;  -1/4   ; 0

amistre64 · Answer

... or other :)

anonymous · Answer

Doesn't look very well bounded to me :P

amistre64 · Answer

Me either :)

amistre64 · Answer

I thought if anything it was the interval between -1 and 1

anonymous · Answer

Which you could say the area is 1/2 total..

amistre64 · Answer

that was my first conjecture as well..

But then I read one source that says odd functions cancel to zero; and others which say to add absolute values of the areas together

anonymous · Answer

You don't need a source to say odd functions cancel to zero :(

amistre64 · Answer

lol..... I wish that was true :)  but im just to much of an idiot at this stage to know the differences ;)

anonymous · Answer

:( If I asked you to find:
$$\int^\pi_{-\pi}x^{10} \sin x \mathbb{d}x $$ would you do it all by parts?

amistre64 · Answer

hmm... I would notice that the graph of the sine is odd and that the part [-pi,0] has the same "area" as [0,pi] and conclude that the total area would be both parts.

anonymous · Answer

:( The integral is 0 because the functions cancel.

HOWEVER, from the exact phrasing of your question, I think you may have to add them together, rather than say 'they cancel', but not completely sure.

amistre64 · Answer

I agree that the functions cancel.  But intuitivly I want to say that the areas combine to total 1/2.  But then the question leaves me to beleieve also that the "bounds" of the function are limitless and not just confined to an interval [-1,1].

The total area under the curve and bounded by the x axis would then be zero to me becasause I can t see trying to take an infinite area....

anonymous · Answer

I think in questions like this it is assumed they mean the 'finite area bounded' - it is just sometimes left out.

amistre64 · Answer

I agree.  But the solutions I put into the answer box all tell me im wrong.  ......

anonymous · Answer

You tried all the solutions in your original post? Hmm

amistre64 · Answer

yep, tried 1/2 to begin with; then figured if it aint that then zero, then the only other options that make sense are 1/4   or -1/4.

I think the programs broke :)

anonymous · Answer

:(

amistre64 · Answer

infinity/2...maybe?  lol

anonymous · Answer

I honestly have no idea what it could be. I agree, it's broken.