evaluate the area of region bounded by the parabola x=y^2 and x=2y-y^2...
It's always when evaluating area of such regions to draw them. I don't like to use the draw feature, so I will try to solve it without drawing. First step you should do here is finding intersection points by setting the two parabolas equal, \[y^2=2y-y^2 \implies 2y^2-2y=0 \implies y=0 \text{ or y=1}.\] This tells us that the two curves will intersect at y=0 and y=1, and those will be the limit of the integral that will give us the area. The integrand should be the difference of the two curves, (the one with higher values inside the interval minus the one with the smaller values.) You can plug y=1/2, for example, to see that \(x=2y-y^2\) is greater inside \((0,1)\), thus the area is \(\large A=\int_0^1 (2y-y^2-y^2)dy \implies A=\int_0^1(2y-2y^2)dy.\)
Join our real-time social learning platform and learn together with your friends!