求定积分∫(0,4)x^2√(4x-x^2)dx
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求定积分∫(0,4)x^2√(4x-x^2)dx
![求定积分∫(0,4)x^2√(4x-x^2)dx](/uploads/image/z/1274962-58-2.jpg?t=%E6%B1%82%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%280%2C4%29x%5E2%E2%88%9A%284x-x%5E2%29dx)
∫[0→4] x²√(4x-x²) dx
=∫[0→4] x²√[4-(x-2)²] dx
令x-2=2sinu,则√[4-(x-2)²]=2cosu,dx=2cosudu,u:-π/2→π/2
=∫[-π/2→π/2] (2sinu+2)²(2cosu)² du
=16∫[-π/2→π/2] (sinu+1)²cos²u du
=16∫[-π/2→π/2] (sin²ucos²u+2sinucos²u+cos²u) du
其中:2sinucos²u为奇函数,其余部分为偶函数
=32∫[0→π/2] (sin²ucos²u+cos²u) du
=8∫[0→π/2] sin²2u du + 16∫[0→π/2] (1+cos2u) du
=4∫[0→π/2] (1-cos4u) du + 16∫[0→π/2] (1+cos2u) du
=4u - sin4u + 16u + 4sin2u |[0→π/2]
=10π
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=∫[0→4] x²√[4-(x-2)²] dx
令x-2=2sinu,则√[4-(x-2)²]=2cosu,dx=2cosudu,u:-π/2→π/2
=∫[-π/2→π/2] (2sinu+2)²(2cosu)² du
=16∫[-π/2→π/2] (sinu+1)²cos²u du
=16∫[-π/2→π/2] (sin²ucos²u+2sinucos²u+cos²u) du
其中:2sinucos²u为奇函数,其余部分为偶函数
=32∫[0→π/2] (sin²ucos²u+cos²u) du
=8∫[0→π/2] sin²2u du + 16∫[0→π/2] (1+cos2u) du
=4∫[0→π/2] (1-cos4u) du + 16∫[0→π/2] (1+cos2u) du
=4u - sin4u + 16u + 4sin2u |[0→π/2]
=10π
若有不懂请追问,如果解决问题请点下面的“选为满意答案”.