Question

How do you evaluate `int (1 + x absx ) dx` for [-2, 3]?

Answer 1

`34/3`

Explanation:

`|x| = x` for `AAx>=0`
`|x| = -x` for `AAx<0`

`I = int_-2^3 (1+x|x|)dx = int_-2^0 (1+x(-x))dx + int_0^3 (1+x*x)dx`

`I = int_-2^0 (1-x^2)dx + int_0^3 (1+x^2)dx`

`I_1 = (x-x^3/3) = 0-0 - (-2) + (-2)^3/3 = 2-8/3 = -2/3`

`I_2 = (x+x^3/3) = 3+3^3/3-0-0=12`

`I = -2/3 + 12 = 34/3`