# How to evaluate the integral of f(x)=sgn x, from x=-1 to x=2? Calculus A Complete Course: Chapter 5.4 Excercise 33 on Page 310. (7th edition by Robert A. Adams & Christopher Essex).

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You need to remember the definition of signum function such that:

`sgn x = {(-1, x<0), (0, x=0), (1, x>0):}`

You need to evaluate the integral `int_-1^2 sgn x dx` , hence, you need to split the integral in two integrals such that:

`int_-1^2 sgn x dx = int_-1^0 sgn x dx + int_0^2 sgn x dx `

`int_-1^2 sgn x dx = int_-1^0 -dx + int_0^2 dx`

Using the fundamental theorem of calculus yields:

`int_-1^2 sgn x dx = -x|_(-1)^0 + x|_0^2`

`int_-1^2 sgn x dx = (0 + (-1)) + 2 - 0`

`int_-1^2 sgn x dx = 2 - 1`

`int_-1^2 sgn x dx = 1`

Hence, evaluating the definite integral of signum function yields `int_-1^2 sgn x dx = 1.`

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