anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Seems like your teacher described everything blow by blow
anonymous
  • anonymous
oh no everything after the second line is me working it....
anonymous
  • anonymous
You wrote a lot, what's your question? (Good work, by the way.)

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anonymous
  • anonymous
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.
anonymous
  • anonymous
Sorry can't follow. What specifically is the problem, integration?
anonymous
  • anonymous
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...
anonymous
  • anonymous
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
anonymous
  • anonymous
\[\int\limits_{0}^{2}(-x ^{2}+2x+3)-3\]
anonymous
  • anonymous
You had a +3, but you have to subtract +3
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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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