Question

How do you find the integral of `abs(x) dx` on the interval [-2, 1]?

Answer 1

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

The easiest way to do this is to think about what this function looks like:

The integral is equal to the area shaded in blue
which is the sum of the areas of the two triangles

`(2xx2)/2 + (1xx1)/2`

`= 2 1/2`

Answer 2

`absx = x` for `x >=0` and `absx = -x` for `x<=0` so to integrate

`int_-2^1 absx dx` we need to split the integral into two integrals:

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

Tlhis method is useful and important to know for more complicated integrals, like: `int_0^8 abs (3x^2-17x+10)dx`,

But for this particular problem there is a straightforward geometric method.

Tlhe graph of `abs x`:

graph{abs x [-3.07, 1.796, -0.313, 2.12]}

From `-2` to `0` we have a right triangle with base = height = 2, so its area is `1/2 2*2 = 2`

On the right, from 0 to 1, is a triangle of area `1/2`, The integrhl is the area below the graph and above the axis, so

`int_-2^1 absx dx = 2+1/2=5/2`

(These are the values of the two integrals above.)