What is the indefinite integral $\int |x-2| dx$?

Question

What is the indefinite integral : $$\int |x-2| dx?$$

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I tried doing it like this : $$\int x-2 dx$$ if $x \geq 2$ $$\int 2-x dx$$ if $x < 2$

$$ \frac {x^2}{2} - 2x + C $$ if $x \geq 2$ $$ 2x - \frac {x^2}{2} +C $$ if $x < 2$

Is that correct ?

Answer 1

Another method is $$\int |x| dx=\dfrac{x|x|}{2}+c\\\vdots \\\int |x-2| dx=\dfrac{(x-2)|x-2|}{2}+c$$

Answer 2

Your approach is correct. Here is another way.

Let $x-2=u\implies \text{d}x=\text{d}u$, hence $$\int |x-2|\text{d}x=\int |u|\text{d}u$$ Now, $$\int |u|\text{d}u=\frac{u}{|u|}\int |u|\cdot \frac{|u|}{u}\text{d}u=\frac{u}{|u|}\int u\text{d}u=\frac{u}{|u|}\cdot \frac{u^2}{2}+C=\frac{u|u|}{2}+C$$Thus $$\int |x-2|\text{d}x=\frac{(x-2)|x-2|}{2}+C$$

Answer 3

For $x>2$, an anti-derivative of $f(x):=|x-2|$ is $\frac{x^2}{2}-2x$; for $x<2$, an anti-derivative of $f$ is $2x-\frac{x^2}{2}$. We can then define $$ F(x)=\begin{cases} \frac{x^2}{2}-2x,&x> 2\\ -(\frac{x^2}{2}-2x),&x<2 \end{cases} $$ so that $F'(x)=f(x)$ for $\mathbb{R}\backslash\{0\}$. If we define $F(2)=0$, then we can check that $F$ is continuous on $\mathbb{R}$. Actually, we have $F$ being differentiable on $\mathbb{R}$ (check it!). In particular, $F'(2)$ exists and is equal to $0=f(2)$. Now we can say that $$ F(x)=\begin{cases} \frac{x^2}{2}-2x,&x\geq 2\\ -(\frac{x^2}{2}-2x),&x<2 \end{cases} $$ gives an anti-derivative of $f$ on $\mathbb{R}$. By the meanvalue theorem, any anti-derivative of $f$ is of the form $F+C$ for some real constant $C$.

Note that one can further write $F$ in a "compact way" as $$ F(x)=\big(\frac{x^2}{2}-2x\big)\hbox{sgn}(x-2). $$ Or alternatively, $$ F(x)=\big(\frac{(x-2)^2}{2}-2\big)\hbox{sgn}(x-2)= \frac{|x-2|(x-2)}{2}-2. $$ Since the constant $-2$ can be "absorbed", one can claim that any anti-derivative of $f$ is of the form $G+C$ where $$ G(x):=\frac{|x-2|(x-2)}{2}. $$