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anonymous

  • 5 years ago

Sketch and find the area bounded by: y=-x^2 + 2x + 3 ; y=3 The graph ends up looking like an upside down paraboloid bound at -2 and +2 with a y intercept at (0,7) integrate: -1/3 x^3 + x^2 + 3x @ x=+2 that equals 22/3 @ x=-2 that equals 2/3 22/3 + 2/3 = 24/3 that would be all the area under the curve from [-2,2] = 8 <--but that seems wrong if I went from [0,2] it would be the 22/3 so my thinking is wrong there somewhere. then the other function integrate y=3 = 3x 3x @ 2 = 6 3x @ -2 = -6 6+6=12 subtract the area in the second function from the first function: 22/3 - 36/3 and wrong answer

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  1. anonymous
    • 5 years ago
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    Seems like your teacher described everything blow by blow

  2. anonymous
    • 5 years ago
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    oh no everything after the second line is me working it....

  3. anonymous
    • 5 years ago
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    You wrote a lot, what's your question? (Good work, by the way.)

  4. anonymous
    • 5 years ago
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    thanks - but somewhere something is wrong. Study guide says answer is 4/3 that above answer is -14/3 but my rational guess is that the answer is more like 8/3 ish.

  5. anonymous
    • 5 years ago
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    Sorry can't follow. What specifically is the problem, integration?

  6. anonymous
    • 5 years ago
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    the line that says @x=-2 that equals 2/3 it really equals -2/3 then when subtracted it become --2/3... I think I have the integration correct (which would be unusual) I'm setting up the problem wrong somehow...

  7. anonymous
    • 5 years ago
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    seems like I should take the definite integral of -x^2 + 2x +3 and then subtract the area under y=3 to get the area above y=3 but apparently not or I'm not doing it right....grrr

  8. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{2}(-x ^{2}+2x+3)-3\]

  9. anonymous
    • 5 years ago
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    You had a +3, but you have to subtract +3

  10. anonymous
    • 5 years ago
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    and that is the right answer....but since the original function is bound by y=3 wouldn't we have to include the same amount of area on the negative side and double the answer or is there some clue in the first two lines to stay positive?

  11. anonymous
    • 5 years ago
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    You are thinking about it too much. The formula is such that you don't have to think about positive area negative area. Along x it is 0 to 2, along y, you have the parabola and subtract y=3.

  12. anonymous
    • 5 years ago
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    haha too much thinking...now I am doomed. Not enough thinking won't help me either I am afraid. Thanks a ton.

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