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How do you evaluate #int (1 + x absx ) dx# for [-2, 3]?

1 Answer

Oct 15, 2015

#34/3#

Explanation:

#|x| = x# for #AAx>=0#
#|x| = -x# for #AAx<0#

#I = int_-2^3 (1+x|x|)dx = int_-2^0 (1+x(-x))dx + int_0^3 (1+x*x)dx#

#I = int_-2^0 (1-x^2)dx + int_0^3 (1+x^2)dx#

#I_1 = (x-x^3/3) = 0-0 - (-2) + (-2)^3/3 = 2-8/3 = -2/3#

#I_2 = (x+x^3/3) = 3+3^3/3-0-0=12#

#I = -2/3 + 12 = 34/3#

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