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_1=x^2 + 5x`
So we need to find:
`=int_-2^0 [(3-x^2)-(x^2+5x)] dx`
`=int_-2^0 [(-2x^2-5x+3)] dx`
`=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.