# integration with absolute value.

#### paulmdrdo

##### Active member
$\displaystyle\int|2x-1|dx$

please tell me what is the first step to solve this.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
$|2x-1|= \left \{ \begin{matrix} 2x-1&\text{ if }&x\geq \dfrac{1}{2}\\1-2x&\text{ if }&x< \dfrac{1}{2}\end{matrix}\right.$

#### paulmdrdo

##### Active member
does that mean i have to integrate those two definitions?

#### Fernando Revilla

##### Well-known member
MHB Math Helper
does that mean i have to integrate those two definitions?
There is only one definition: a piecewise function with two subdomains. Integrate at each subdomain.

#### paulmdrdo

##### Active member
we will have two answers is this correct?
$\displaystyle \int (2x-1)dx = {x^2}+x+C\,\,if\,x\geq\frac{1}{2}$
$\displaystyle \int (1-2x)dx = x-{x^2}+C\,\,if\,x\leq\frac{1}{2}$

#### Fernando Revilla

##### Well-known member
MHB Math Helper
$\displaystyle \int (2x-1)dx = {x^2}+x+C\,\,if\,x\geq\frac{1}{2}$
$\displaystyle \int (1-2x)dx = x-{x^2}+C\,\,if\,x\leq\frac{1}{2}$
Check the first one.

#### paulmdrdo

##### Active member
oh i made a typo. it should be,
$\displaystyle \int (2x-1)dx = {x^2}-x+C\,\,if\,x\geq\frac{1}{2}.$

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follow-up question, what if it is bounded?

$\displaystyle\int_{-1}^1|2x-1|dx$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
we will have two answers is this correct?
$\displaystyle \int (2x-1)dx = {x^2}+x+C\,\,if\,x\geq\frac{1}{2}$
$\displaystyle \int (1-2x)dx = x-{x^2}+C\,\,if\,x\leq\frac{1}{2}$
Besides $x^2-x$ in the first formula, an indefinite integral, by definition, is a differentiable, and therefore a continuous, function. So, the constants in the two lines have to be such that at 1/2 the functions have the same value.

follow-up question, what if it is bounded?

$\displaystyle\int_{-1}^1|2x-1|dx$
You still compute two definite integrals: from -1 to 1/2 plus from 1/2 to 1.

#### paulmdrdo

##### Active member
evgenymakarov "So, the constants in the two lines have to be such that at 1/2 the functions have the same value." - what do you mean by this?

and why is it from-1 to 1/2 plus 1/2 to 1?

#### Plato

##### Well-known member
MHB Math Helper
follow-up question, what if it is bounded?
$\displaystyle\int_{-1}^1|2x-1|dx$
why is it from-1 to 1/2 plus 1/2 to 1?
Look at the plot here.

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#### paulmdrdo

##### Active member
can you show me your work? i'm kind of confused because the boundary is from -1 to 1 only. why did you insert 1/2 there?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
evgenymakarov "So, the constants in the two lines have to be such that at 1/2 the functions have the same value." - what do you mean by this?
It's not like there are many lines with formulas and constants in the first quote in post #8. Anyway, to understand this I suggest you write explicitly the answer to your original problem in post #1, i.e., the result of evaluating this indefinite integral. And remember that the resulting function should be differentiable and, in particular, continuous. .

and why is it from-1 to 1/2 plus 1/2 to 1?
Try evaluating it any other way algebraically. Because the function being integrated is a piecewise function that is given by simple but different algebraic expressions below 1/2 and above 1/2, as shown in post #2.

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#### paulmdrdo

##### Active member
what i have in my mind is like this

$\displaystyle\int_{-1}^1 (2x-1)dx$
$\displaystyle\int_{-1}^1 (1-2x)dx$

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Besides $x^2-x$ in the first formula, an indefinite integral, by definition, is a differentiable, and therefore a continuous, function. So, the constants in the two lines have to be such that at 1/2 the functions have the same value.
Right. And considering it, we can express the solution in a more elegant way:
$$\int |2x-1|\;dx=\frac{(2x-1)\left |2x-1\right|}{4}+C$$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
what i have in my mind is like this

$\displaystyle\int_{-1}^1 (2x-1)dx$
$\displaystyle\int_{-1}^1 (1-2x)dx$
I assume this is related to evaluating $\int_{-1}^1|2x-1|\,dx$, but I am not sure how it is related.

#### paulmdrdo

##### Active member
$\displaystyle\int |2x-1|\;dx=\frac{(2x-1)\left |2x-1\right|}{4}+C$ --- why the other (2x-1) is in absolute value?

#### Fernando Revilla

##### Well-known member
MHB Math Helper
You can easily verify that if
$$f(x)= \left \{ \begin{matrix} \;\;\;x^2-x+\dfrac{1}{4}&\text{if}&x\geq \dfrac{1}{2}\\ -x^2+x-\dfrac{1}{4}&\text{if}&x< \dfrac{1}{2}\end{matrix}\right.$$
then, $f'(x)=\left|2x-1\right|$ for all $x\in\mathbb{R}$. But
$$\frac{(2x-1)|2x-1|}{4}=\left \{ \begin{matrix} \dfrac{(2x-1)(2x-1)}{4}&\text{if}&x\geq \dfrac{1}{2}\\ \frac{(2x-1)(1-2x)}{4}&\text{if}&x< \dfrac{1}{2}\end{matrix}\right.\\= \left \{ \begin{matrix} \;\;\;x^2-x+\dfrac{1}{4}&\text{if}&x\geq \dfrac{1}{2}\\ -x^2+x-\dfrac{1}{4}&\text{if}&x< \dfrac{1}{2}\end{matrix}\right.$$
As a consequence,
$$\int |2x-1|\;dx=\frac{(2x-1)\left |2x-1\right|}{4}+C\quad (C\in\mathbb{R})$$

#### paulmdrdo

##### Active member
oh men, i'm lost here!

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
As Fernando pointed out, writing the result of integration using absolute value is more elegant, but not necessarily easier.

The idea is simple. You have a piecewise function that is made of two functions: before 1/2 it is $1 - 2x$ and after 1/2 it is $2x - 1$. Correspondingly, the indefinite integral will also be different before and after 1/2. You just need to integrate the corresponding function. The definite integral can also be broken into two parts (before and after 1/2) because in each part the function being integrated is either $1-2x$ or $2x-1$, and integrating such functions is straigtforward.