amistre64
  • amistre64
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
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
... or other :)
anonymous
  • anonymous
Doesn't look very well bounded to me :P
amistre64
  • amistre64
Me either :)

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amistre64
  • amistre64
I thought if anything it was the interval between -1 and 1
anonymous
  • anonymous
Which you could say the area is 1/2 total..
amistre64
  • amistre64
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
  • anonymous
You don't need a source to say odd functions cancel to zero :(
amistre64
  • amistre64
lol..... I wish that was true :) but im just to much of an idiot at this stage to know the differences ;)
anonymous
  • anonymous
:( If I asked you to find: \[\int^\pi_{-\pi}x^{10} \sin x \mathbb{d}x \] would you do it all by parts?
amistre64
  • amistre64
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
  • anonymous
:( 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
  • amistre64
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
  • anonymous
I think in questions like this it is assumed they mean the 'finite area bounded' - it is just sometimes left out.
amistre64
  • amistre64
I agree. But the solutions I put into the answer box all tell me im wrong. ......
anonymous
  • anonymous
You tried all the solutions in your original post? Hmm
amistre64
  • amistre64
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
  • anonymous
:(
amistre64
  • amistre64
infinity/2...maybe? lol
anonymous
  • anonymous
I honestly have no idea what it could be. I agree, it's broken.

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