Matematika

Pertanyaan

hasil integral 0-1 (3x^2-16x-12)dx

1 Jawaban

  • 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

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