Question

What is `int_(-4)^(1) |x^2-4|dx`?

Answer 1

`int_(-4)^1 abs(x^2-4)dx = 59/3`

Explanation:

Consider the function:

`f(x) = x^2-4 = (x+2)(x-2)`

we know that a second polynomial with positive leading coefficient is negative in the interval between its roots and negative outside the interval, so:

`f(x) > 0` for `x in(-oo,-2) uu (2,+oo)`

`f(x) < 0` for `x in (-2,2)`

Using the additivity of the definite integral we can split it as:

`int_(-4)^1 abs(x^2-4)dx = int_(-4)^-2abs(x^2-4)dx + int_(-2)^1 abs(x^2-4)dx`

`int_(-4)^1 abs(x^2-4)dx = int_(-4)^-2 (x^2-4)dx - int_(-2)^1 (x^2-4)dx`

Now:

`int_(-4)^-2 (x^2-4)dx = [x^3/3-4x]_(-4)^(-2) = -8/3+8+64/3-16 = 32/3`

`int_(-2)^1 (x^2-4)dx = [x^3/3-4x]_(-2)^1 = 1/3-4+8/3-8 = -9`

So:

`int_(-4)^1 abs(x^2-4)dx = 32/3+9 = 59/3`