hasil integral 0-1 (3x^2-16x-12)dx
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Pertanyaan
hasil integral 0-1 (3x^2-16x-12)dx
1 Jawaban
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1. Jawaban ajengchelsania
int 0 -1 (3x^2 - 16x - 12)dx
= int (3x^2 - 16x - 12)dx
= 3/(2+1) x^(2+1) - 16/(1+1) x^(1+1) - 12x
= 3/3 x^3 - 16/2 x^2 - 12x
= x^3 - 8x^2 - 12x ]0 -1
= [0^3 - 8(0)^2 - 12(0)] - [(-1)^3 - 8(-1)^2 -12(-1)]
= [0] - [-1 - 8 + 12]
= [0] - [3]
= -3
*semoga membantu