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Misal u =sin(x) du = cos(x) dx so, = int [0,π/2] u² du = [1/3 u³] [0,π/2] = [1/3 sin³x] [0,π/2] = (1/3 sin³(π/2)) - 1/3 sin³(0) = 1/3 (1³) - 1/3 (0) = 1/3
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MathSolver74
koreksi
itu batas [0 , π/2] utk variabel x
jika kamu misalin u maka batasan berubah
x = 0 >> sin (0) = 0
x = π/2 >> sin (π/2) = 1
btsnya jd [0 , 1]
du = cos(x) dx
so,
= int [0,π/2] u² du
= [1/3 u³] [0,π/2]
= [1/3 sin³x] [0,π/2]
= (1/3 sin³(π/2)) - 1/3 sin³(0)
= 1/3 (1³) - 1/3 (0)
= 1/3