mohon bantuannya ya teman2, sekalian caranya juga
Matematika
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Pertanyaan
mohon bantuannya ya teman2, sekalian caranya juga
1 Jawaban
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1. Jawaban ahreumlim
15 . ∫(4 cos2x - 3 sin3x)dx (dari π/3 sampai π/2)
= 4/2 sin2x + 3/3 cos3x
= 2 sin2x + cos3x (dievaluasi dari π/3 sampai π/2)
= 2 sin(2π/2) + cos(3π/2) - 2 sin(2π/3) - cos(3π/3)
= 2 sin(π) + cos(3π/2) - 2 sin(2π/3) - cos(π)
= 0 + 0 - 2 √3/2 - (-1)
= 1 - √3 ==>> A
16. ∫(sin3x + cosx) dx (dari 0 samai π)
= -(1/3) cos3x + sinx (di evaluasi dari 0 samai π)
= -(1/3) cos(3π) + sin(π) + (1/3) cos(0) - sin(0)
= -(1/3) (-1) + 0 + (1/3) - 0
= (1/3) + (1/3)
= 2/3 ==>> D
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