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:
`=int_-3^0.5 ([3-x^2]-[x^2+5x]) dx`
`=14.29\ text[sq units]`