find the area of the regin bounded by the graphs of y=-x^2=2x=3 and y=3.
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