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Question:

Evaluate: ∫31[|x𕒵|+|x𕒶|+|x𕒷|]dx

Solution:

First calculate indefinite integral:
I=∫[|x𕒷|+|x𕒶|+|x𕒵|]dx=∫[1/2(x𕒷)|x𕒷|+1/2(x𕒶)|x𕒶|+1/2(x𕒵)|x𕒵|]dx
By the Fundamental Theorem of Calculus ∫baF(x)dx=f(b)−f(a), so just evaluate integral at endpoints.
[1/2(x𕒷)|x𕒷|+1/2(x𕒶)|x𕒶|+1/2(x𕒵)|x𕒵|]31dx=5/2+5/2=5