In calculus has two parts. There are differentiation and integration. Integration is defined as the process of reverse differentiation.

If the function is f(x). The integration of f(x) can be expressed as `int ` f(x) dx. where, f(x) is continues function. The definite integrals has upper limit and lower limits. It can be
expressed as `int_a^b` f(x) dx. Absolute value is always positive value. For example | -8 + 4 | = 4 . In this article, we shall discuss about integral of abs x.

Integral Problems:

Integral problem 1:

What is the integral of abs x with an interval 0 to 2.

Solution:

The integral absolute function is `int_0^2` | x | dx .

= `[ | x^2 / 2 | ] _ 0^2` .

Apply the upper and lower limits.

= ` [ | 2^2/2 | ] - [ | 0/2| ]` .

= ` [ | 4/2| ] ` .

= | 2 | .

= 2 .

Answer: The integral of abs x is 2 .

Integral Problem 2:

What is the integral of abs x with an interval -2 to 0.

Solution:

The integral absolute function is `int_(-2)^0` | x | dx . = `int_0^2` | x | dx .

= `[ | x^2 / 2 | ] _ 0^2` .

Apply the upper and lower limits.

= ` [ | 2^2/2 | ] - [ | 0/2| ]` .

= ` [ | 4/2| ] ` .

= | 2 | .

= 2 .

Answer: The integral of abs x is 2 .

Integral problem 3:

What is the integral of abs (x - 1) with an interval 0 to 5.

Solution:

The integral absolute function is = `int_0^5` | x - 1 | dx .

= `[ | (x^2 / 2) - x | ] _ 0^5` .

Apply the upper and lower limits.

= ` [ | (5^2/2) - 5 | ] - [ | (0/2) - 0 | ]` .

= ` [ | (25/2) - 5| ] ` .

= ` | (25 - 10)/2 |` .

= ` |15/2|` .

Answer: The integral of abs (x - 1) is ` |15/2|` .

Integral problem 4:

What is the integral of abs (x - 3) with an interval 0 to 5.

Solution:

The integral absolute function is = `int_0^5` | x - 3 | dx .

= `[ | (x^2 / 2) - 3x | ] _ 0^5` .

Apply the upper and lower limits.

= ` [ | (5^2/2) - 3(5) | ] - [ | (0/2) - 3(0) | ]` .

= ` [ | (25/2) - 15| ] ` .

= ` | (25 - 15)/2 |` .

= ` |10/2|` .

= 5 .

Answer: The integral of abs (x - 3) is 5 . .

Integral problem 5:

What is the integral of abs (-x) with an interval 0 to 1.

Solution:

The integral absolute function is `int_0^1` | - x | dx .

= `[ | - x^2 / 2 | ] _ 0^1` .

Apply the above upper and lower limits.

= ` [ | - 1^2/2 | ] - [ | 0/2| ]` .

= ` [ | - 1/2| ] ` .

= `(1 / 2 )` .

Answer: The integral of abs (-x) is `(1/2)` .