Sketch first:

1 2 3 -1 -2 -3 -4 -5 -6 5 10 -5 -10 x y
`y=x^2 + 5x`
`y=3-x^2`

Area bounded by the curves `y=x^2 + 5x` and `y=3-x^2`, including a typical rectangle.

We need to use: `A=int_a^b(y_2-y_1) dx`

We note that `y = 3 − x^2` is above `y = x^2+ 5x` so we take

`y_2= 3 − x^2` and `y_1= x^2+ 5x`

Points of intersection occur where:

`x^2 + 5x = 3 − x^2`

`2x^2 + 5x − 3 = 0`

`(x + 3)(2x − 1) = 0`

So `x = -3` or `x = 0.5`

We take vertical elements (indicated by the vertical rectangle in the graph above).

So the area is given by:

`text[Area]=int_a^b(y_2-y_1) dx`

`=int_-3^0.5 ([3-x^2]-[x^2+5x]) dx`

`=int_-3^0.5(3-5x-2x^2) dx`

`=[3x-(5x^2)/(2)-(2x^3)/(3)]_-3^0.5`

`=14.29\ text[sq units]`