Intergral 1 sd -1 ( 2x^2 - 4x + 3 ) dx
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Intergral 1 sd -1 ( 2x^2 - 4x + 3 ) dx
1 Jawaban
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1. Jawaban imronmaulana212
= [2/3 x^3 - 2x^2 + 3x] -1 sd 1
=[ (2/3 (1) - 2(1) + 3(1) ] - [2/3(-1)^3 - 2(-1)^2 + 3(-1)
= (2/3) -2 + 3 - [-2/3 - 2 -3]
= 4/3 +6
= 22/3