Question

Nilai dari adalah ?

Answer 1

Int [0 - π/3] sin x cos x dx
= int [0 - π/3] (1/2) sin 2x dx
= (1/2) [(-1/2) cos 2x] [0 - π/3]
= (1/2) [(-1/2)(-1/2) - (-1/2)(1)]
= (1/2) [(1/4 + 1/2)]
= (1/2)(3/4)
= 3/8

Answer 2

2SinxCosx = Sin2x ⇔ SinxCosx = (1/2)Sin2x
Int batas bawah 0 batas atas π/3 dari SinxCosx dx
= Int batas bawah 0 batas atas π/3 dari (1/2)Sin2x dx
= -(1/4)Cos 2x ] batas bawah 0 batas atas π/3
= (-1/4)(Cos2π/3 - Cos 0)
= (-1/4)(-1/2 - 1)
= (-1/4)(-3/2)
= 3/8