Question

How do you solve `int_4^-2 |x-1|dx`?

Answer 1

`-9`

Explanation:

Firstly, you evaluate rather than solve ;-)

But moving on to the real stuff, we just need to break that absolute vaue thing out

`|x-1| =-( x-1), x in (-oo,1]`
and
`|x-1| = x-1, x in [1, oo)`

so first we split the integral

`int_4^-2 |x-1|dx`

`= int_4^1 |x-1| \ dx + int_1^-2 |x-1| \ dx`

`= int_4^1 x-1 \ dx + int_1^-2 -(x-1) \ dx`

switching the limits on the second integral
`= int_4^1 x-1 \ dx + int_-2^1 x-1 \ dx`

`= [ x^2 / 2 - x]\_4^1 + [ x^2/2-x ]_(-2)^1`

`= 2 [ x^2 / 2 - x]\_(x=1) - [ x^2/2-x ]_(x=4) - [ x^2/2-x ]\_(x=-2)`

`= 2 [- 1/2] - [4 ] - [ 4 ] = -9`

We could also have recognised the symmetry, namely that the

`int_4^-2 |x-1|dx = 2 int_4^1 x-1 \ dx`

which becomes apparent from sketching the function itself