Ответ(ы) на вопрос:
Гость 7
∫(√(49 -x²)dx)
0
замена: x =7sint,0=7sint⇒ t₁=0 ;7=7sint ⇒t₂=π/2 ;dx =7costdt ;t₂=π/2 ||
π/2 π/2 π/2
∫(7cost*7costdt) =49∫(1+cos2t)/2 dt =(49/2)(t +(1/2)sin2t) |
0........................... 0......................................................0
= (49/2)*(π/2 +(1/2)sin2*π/2 - 0 -(1/2)sin2*0) =
=(49/2)*(π/2 +(1/2)sinπ - 0 -(1/2)sin0) =(49/2)*(π/2 +0 - 0 -0)
= 49π/4.
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