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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
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Seems like your teacher described everything blow by blow

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh no everything after the second line is me working it....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You wrote a lot, what's your question? (Good work, by the way.)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks  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
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry can't follow. What specifically is the problem, integration?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0seems 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{2}(x ^{2}+2x+3)3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You had a +3, but you have to subtract +3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0haha 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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