The integral $$\int_0^2 ||x - 1| - x| \; dx$$ is equal to ___________.

We want to evaluate the definite integral

$$I=\int_{0}^{2}\bigl\lvert\,\lvert x-1\rvert-x\,\bigr\rvert\,dx.$$

First we simplify the integrand step by step. We begin with the inner absolute value. For any real $$x$$ we have the basic property $$|a-b|=\left\{\begin{array}{ll}a-b,&a\ge b\\b-a,&a\lt b\end{array}\right.$$ and we will apply this repeatedly.

For $$|x-1|$$ we examine the sign of $$(x-1)$$ on the interval $$[0,2]$$.

• When $$0\le x\le1$$, $$(x-1)\le0$$, so $$|x-1|=1-x.$$

• When $$1\le x\le2$$, $$(x-1)\ge0$$, so $$|x-1|=x-1.$$

Substituting this into the integrand we introduce the function

$$f(x)=\bigl|\,|x-1|-x\,\bigr|.$$

Now we analyse $$f(x)$$ piece-wise.

Case 1 : $$0\le x\le1$$

Here $$|x-1|=1-x,$$ so

$$|x-1|-x=(1-x)-x=1-2x.$$

We must now take the absolute value of $$(1-2x)$$, i.e. evaluate $$|1-2x|.$$ The sign of $$(1-2x)$$ changes at the point where $$1-2x=0\implies x=\tfrac12$$. Hence we split the subinterval $$[0,1]$$ a second time.

• When $$0\le x\le\frac12$$, we have $$1-2x\ge0$$, so $$|1-2x|=1-2x.$$

• When $$\frac12\le x\le1$$, we have $$1-2x\le0$$, so $$|1-2x|=-(1-2x)=2x-1.$$

Case 2 : $$1\le x\le2$$

Here $$|x-1|=x-1,$$ so

$$|x-1|-x=(x-1)-x=-1.$$

The absolute value of a constant $$-1$$ is simply $$|-1|=1.$$ Thus on the entire interval $$[1,2]$$ the function $$f(x)$$ equals $$1$$.

Gathering all this information, we have the piece-wise definition

$$ f(x)=\begin{cases} 1-2x,&0\le x\le\frac12,\\[4pt] 2x-1,&\frac12\le x\le1,\\[4pt] 1,&1\le x\le2. \end{cases} $$

With the integrand now expressed explicitly, we split the original integral into three simpler integrals:

$$ I=\int_{0}^{\frac12}(1-2x)\,dx+\int_{\frac12}^{1}(2x-1)\,dx+\int_{1}^{2}1\,dx. $$

We evaluate each integral individually.

First integral

Using the power rule $$\displaystyle\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$$ (where $$n\neq-1$$), we integrate:

$$ \int_{0}^{\frac12}(1-2x)\,dx =\left[x-\frac{2x^{2}}{2}\right]_{0}^{\frac12} =\left[x-x^{2}\right]_{0}^{\frac12}. $$

Substituting the limits:

$$ \bigl(\tfrac12-(\tfrac12)^2\bigr)-\bigl(0-0^2\bigr) =\left(\tfrac12-\tfrac14\right)-0 =\tfrac14. $$

Second integral

Again applying the power rule,

$$ \int_{\frac12}^{1}(2x-1)\,dx =\left[x^{2}-x\right]_{\frac12}^{1}. $$

Substituting the limits:

$$ \bigl(1^{2}-1\bigr)-\bigl((\tfrac12)^{2}-\tfrac12\bigr) =(1-1)-\left(\tfrac14-\tfrac12\right) =0-\left(-\tfrac14\right) =\tfrac14. $$

Third integral

This one is immediate because the integrand is the constant $$1$$:

$$ \int_{1}^{2}1\,dx =[x]_{1}^{2}=2-1=1. $$

Finally we add the three results:

$$ I=\tfrac14+\tfrac14+1=\tfrac12+1=\tfrac32=1.5. $$

So, the answer is $$1.5$$.