Computing the integral $\int_{-1}^{2} \left( |x| + |1-x| \right) {\rm d} x$

Question

I'm having trouble with one of the exercises, I have to split the integral for the absolute value but I can't manage to algebraic find the boundaries for the integral.

$$\int_{-1}^{2} \left( |x| + |1-x| \right) {\rm d} x$$

Thanks in advance!

Answer 1

$|x|=x$ when $x>0$ and $|x|=-x$ when $x<0$. Similarly, $|1-x|=1-x$ when $1-x>0$ and $|1-x|=x-1$ when $1-x<0$. We end up with three different cases: $x<0$, $0<x<1$, and $x>1$. For example, when $x<0$, we have

$$|x|+|1-x|=-x+1-x=1-2x$$

which you should have no problem integrating now that the absolute values are gone. You should be able to handle the other two cases.

Answer 2

$$\int_{-1}^2 dx (|x|+|1-x|) = \int_{-1}^0 dx (-x + 1-x) + \int_0^1 dx (x+1-x) + \int_1^2 dx [x-(1-x)] $$

Go...

Answer 3

There are three regions:

$$\int_{-1}^{2} (|x|+|1-x|) dx=\int_{-1}^{0}-2x+1 \ dx +\int_{0}^{1}1\ dx+\int_{1}^{2}2x-1 \ dx=5.$$

To verify this, take a look at the following figure:

Answer 4

The point where $|x|$ 'changes form' (between $x$ and $-x$) is when $x=0$.

The point where $|1-x|$ changes form is when $1-x = 0$

Answer 5

You don't need to do the actual integration, just remember that you're summing the area under the curves. Notice that $$\int_{-1}^{2} (|x|+|1-x|) dx$$ equals $$\int_{-1}^{2} (|x|) dx + \int_{-1}^{2} (|1-x|) dx$$ Then just draw both parts on a graph. You'll see that each part specifies two triangles: one with an area of 0.5 and the other with an area of 2. Add that up $(2 + 0.5 + 2 + 0.5)$ to get 5