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

Question

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 · Answer

Seems like your teacher described everything blow by blow

anonymous · Answer

oh no everything after the second line is me working it....

anonymous · Answer

You wrote a lot, what's your question? (Good work, by the way.)

anonymous · Answer

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 · Answer

Sorry can't follow. What specifically is the problem, integration?

anonymous · Answer

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 · Answer

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 · Answer

$$\int\limits_{0}^{2}(-x ^{2}+2x+3)-3$$

anonymous · Answer

You had a +3, but you have to subtract +3

anonymous · Answer

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 · Answer

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 · Answer

haha too much thinking...now I am doomed.  Not enough thinking won't help me either I am afraid.   Thanks a ton.