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December 22, 2014

December 22, 2014

Posted by **david** on Monday, January 8, 2007 at 8:31pm.

i don't need help solving the problem and but i am a little confused.

ok the graph is a parabola and i drew a parobla with y= 3. now when find the area, am I finding the area on the top, above y=3 or on the bottom, below y=3.

i hope my question makes sense. basically, whick part of the graph am I finding the area of. top or bottom?

sorry about the equation. it is actually y=-x^2+2x+3

find the area from the top of the parabola down to y=3

above ("top" of) y=3

y=3 cuts the parabola at (-2,3)and (0,3), so the area bounded is the area of the parabola below the line y=3.

its area is integral(3 - x^2 - 2x - 3)dx from -2 to 0

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