# 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 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 Q1: What is the modulus of x? Ans: The modulus of x is defined as follows: |x|=x if x>0, and |x|=-x if x$\leq$0. Q2: What is the integration of |x|? Ans: The integration of mod x is$-\frac{x|x|}{2}+c\$.

Q3: Is mod x differentiable at x=0?

Ans: The mod x is not differentiable at x=0.