Question

How do you evaluate the integral of `int absx dx` from -2 to 1?

Answer 1

Split the interval into two.

Explanation:

`int_-2^1 absx dx = int_-2^0 absx dx + int_0^1 absx dx`

For `x<0`, `absx=-x`.
For `x>=0`, `absx=x`.

`int_-2^0 absx dx + int_0^1 absx dx = int_-2^0 -x dx + int_0^1 x dx`

`= [-x^2/2]_-2^0 + [x^2/2]_0^1`

`= 2 + 1/2`

`= 5/2`

Answer 2

`int_(-2)^1|x|dx = 5/2`

Explanation:

The absolute value function `|*|` can be written as the piecewise function
`|x| = {(-x" if "x<=0), (x" if "x>=0):}`

Thus, using the property that for `a < b < c`
`int_a^cf(x)dx = int_a^bf(x)dx + int_b^cf(x)dx`

we obtain

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

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

`=-int_-2^0xdx + int_0^1xdx`

`= -[x^2/2]_-2^0 + [x^2/2]_0^1`

`=-(0^2/2 - (-2)^2/2) + (1^2/2 - 0^2/2)`

`= -(-2)+1/2`

`=5/2`

Thus `int_(-2)^1|x|dx = 5/2`