Matematika

Pertanyaan

Tolong yang pintar integral mohon dijawab ya.. besok dikumpul..
Tolong yang pintar integral mohon dijawab ya.. besok dikumpul..

1 Jawaban

  • No 28
    ∫ 3 cos 2x dx = 3 ∫ cos 2x dx
    = 3 × 1/2 sin 2x
    = 3/2 sin 2x
    Masukkan batas integralnya:
    3/2 ×[ sin2(π/4) - 3/2 × sin 2(π/12)\
    = 3/2[sin π/2 - sin π/6 ]
    = 3/2[1 - 1/2]
    = 3/2(1/2)
    = 3/4 (B)

    No 29
    ∫2 sin²x dx = 2 ∫ sin²x dx
    =2 × ∫(1-cos 2x)/2 dx
    = 2 × 1/2 × ∫1 - cos 2x dx
    = ∫1 - cos 2x dx
    = x - 1/2 sin 2x dx
    Masukkan batas integralnya
    =(π/2 - 1/2 sin 2(π/2) ) - (0 - 1/2 sin 2(0))
    = (π/2 - 1/2 × 0) - (0 - 0)
    = π/2 - 0 - 0
    = π/2 = 1/2 π (C)