Discover Common Mistakes: What Am I Doing Wrong?

[camboguy]

what am i doing wrong?

 

[Mark44]

You have the right value for the integral you show. If you are calculating area, then you shouldn't get a negative value, so your integral is set up wrong in that case.

 

[HallsofIvy]

If the problem was "Find the area between $x= y^2- 4y$ and $x= 2y- y^2$ for y between 0 and 3" you should note that $2y- y^2> y^2- 4y$ for y between 0 and 3 so you have the subtraction backwards.