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amistre64

  • 5 years ago

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

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  1. amistre64
    • 5 years ago
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    ... or other :)

  2. anonymous
    • 5 years ago
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    Doesn't look very well bounded to me :P

  3. amistre64
    • 5 years ago
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    Me either :)

  4. amistre64
    • 5 years ago
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    I thought if anything it was the interval between -1 and 1

  5. anonymous
    • 5 years ago
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    Which you could say the area is 1/2 total..

  6. amistre64
    • 5 years ago
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    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

  7. anonymous
    • 5 years ago
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    You don't need a source to say odd functions cancel to zero :(

  8. amistre64
    • 5 years ago
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    lol..... I wish that was true :) but im just to much of an idiot at this stage to know the differences ;)

  9. anonymous
    • 5 years ago
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    :( If I asked you to find: \[\int^\pi_{-\pi}x^{10} \sin x \mathbb{d}x \] would you do it all by parts?

  10. amistre64
    • 5 years ago
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    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.

  11. anonymous
    • 5 years ago
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    :( 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.

  12. amistre64
    • 5 years ago
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    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....

  13. anonymous
    • 5 years ago
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    I think in questions like this it is assumed they mean the 'finite area bounded' - it is just sometimes left out.

  14. amistre64
    • 5 years ago
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    I agree. But the solutions I put into the answer box all tell me im wrong. ......

  15. anonymous
    • 5 years ago
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    You tried all the solutions in your original post? Hmm

  16. amistre64
    • 5 years ago
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    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 :)

  17. anonymous
    • 5 years ago
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    :(

  18. amistre64
    • 5 years ago
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    infinity/2...maybe? lol

  19. anonymous
    • 5 years ago
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    I honestly have no idea what it could be. I agree, it's broken.

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