Sketching first:

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

Graphs of `y=x^2 + 5x` and `y=3-x^2`, showing the portion between `-2 < x < 0`.

From the graph, we see that `y=3-x^2` is above `y=x^2 + 5x` in the region of interest, so we'll use:

`y_2=3-x^2`, and

`y_1=x^2 + 5x`

So we need to find:

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

`=int_-2^0 [(3-x^2)-(x^2+5x)] dx`

`=int_-2^0 [(-2x^2-5x+3)] dx`

`=[-2/3x^3-5/2x^2+3x]_-2^0`

`=0-[16/3-10-6]`

`=10 2/3\ text(sq units)`

Some of the shaded area is above the `x`-axis and some of it is below. Don't worry about taking absolute value - the formula takes care of that automatically.