Sketch first:

1 2 3 -1 -2 -3 5 x y
`y=x^2 + 5x`
`y=3-x^2`

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`.

`text[Area]=int_c^d(x_2-x_1) dy`

`=int_1^4([2-y]-[-sqrt[y]]) dy`

`=int_1^4(2-y+sqrt[y])dy`

`=[2y-(y^2)/(2)+2/3 y^(3//2)]_1^4`

`=(16/3)-(13/6)`

`=19/6\ text[sq units]`