Area bounded by `y = x^2`, `y = 2 − x` and `y = 1`, including a typical rectangle.
We take horizontal elements in this case.
So we need to solve `y = x^2` for `x`:
`x = ±sqrt(y)`
We need the left hand portion, so `x = − sqrt(y)`.
Notice that `x = 2 − y` is to the right of `x = -sqrt(y)` so we choose
`x_2= 2 − y` and `x_1= -sqrt(y.)`
The intersection of the graphs occurs at `(-2,4)` and `(1,1)`.
So we have: `c = 1` and `d = 4`.
`=19/6\ text[sq units]`