Integration of mod x | Integral of modulus x

Integration of |x|. The integration of mod x is equal to ∫|x| dx = (-x|x|)/2 +C where C is an arbitrary constant. Before we find the integration of modulus of x or the integration of the absolute value of x, we need to know what |x| is. Then we will calculate the integration of mod x.

What is mod x?

If $x$ is a real number number, then $|x|$ is defined as follows:

$|x|=x \quad$ if $ x \geq 0 \\$ $\quad \,\,=-x \,\,$ if $x<0$

So $| \,|$ always takes positive values. From the definition, it is clear that $| \,|:\mathbb{R} \to \mathbb{R}^+$ is a surjective function but not injective.

Now, we will find the integration of the modulus of x.

What is Integration of mod x?

Question: Evaluate $\int |x| dx$

Solution:

Case 1: First we assume that $x \geq 0$

So $|x|=x$

$\therefore \int |x| dx=\int x dx$

$=\dfrac{x^2}{2}+c$

$\big[\because x^n dx=\dfrac{x^{n+1}}{n+1}+c\big]$

$=\dfrac{1}{2}x^2+c$

Case 2: Next we assume that $x<0$

So $|x|=-x$

$\therefore \int |x| dx=\int (-x) dx$

$=-\int x dx$

$=-\dfrac{x^2}{2}+c$

$=-\dfrac{1}{2}x^2+c$

∴ From case 1 and case 2, we obtain that

$\int |x| dx=\frac{1}{2}x^2+c \quad$ if $x \geq 0$
$\quad \quad \quad \,=-\frac{1}{2}x^2+c \,\,$ if $x<0$

Combining both the above cases, we deduce that the integration of absolute x is

\[\int |x| dx=-\dfrac{x|x|}{2}+c\]
where $c$ is an integration constant

So the integration of |x| is equal to ∫|x| dx = (-x|x|)/2 +C where C is an integral constant.

Read Also: Derivative of mod x

Generalisation of mod x: For any real number $a$, we have

$|x-a|=x-a \quad \quad$ if $x \geq a$ $\quad \quad \,\,=-(x-a) \,\,$ if $x<a$

Thus, using the same method as above we can show that

\[\int |x-a| dx=-\dfrac{(x-a)|x-a|}{2}+c\]

Also Read: 

Derivative of root x: The derivative of √x is 1/2√x

Integration of root x: The integration of  √x is 2/3x^{3/2}

Derivative of cube root of x: The derivative of the cube root of x is 1/(3x^{2/3})

Now we will find the definite integral of mod x from -1 to 1.

Integral of |x| from -1 to 1.

Question: Evaluate $\int_{-1}^1 |x| dx$

Solution:

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

$=\int_{-1}^0 |x| dx+\int_0^1 |x| dx$

[We know that |x|=-x if -1<x<0 and |x|=x if 0<x<1]

$=\int_{-1}^0 (-x) dx+\int_0^1 x dx$

$=-\int_{-1}^0 x dx+\int_0^1 x dx$

$=-\Big[\dfrac{x^2}{2} \Big]_{-1}^0+\Big[\dfrac{x^2}{2} \Big]_0^1$

$\big[\because x^n dx=\dfrac{x^{n+1}}{n+1}\big]$

$=-[\frac{0^2}{2}-\frac{(-1)^2}{2}]+[\frac{1^2}{2}-\frac{0^2}{2}]$

$=-[0-\frac{1}{2}]+[\frac{1}{2}-0]$

$=\frac{1}{2}+\frac{1}{2}$

= 1.

So the integral of |x| from -1 to 1 is equal to 1.

FAQs of Integration of mod x

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