The solution to is x = 1 and x = -2.Hello,
I have been thinking about this double integral for a couple hours and I can't seem to recall how to properly set it up.
I am attempting to integrate something under the surface of z = 1 / (y+2) and over the area which is bounded by y=x and y^2+x=2.
I initially figured that this integral would be simpler if you integrated with respect to 'x' first. Making the lower bound for x, y; and the upper bound 2-y^2. From this I got the y lower boundary to be -sqrt 2 and upper bound sqrt 2. I have attempted multiple variations and cannot seem to arrive at the correct answer.
Can anyone assist me?