高数,定积分。我怎么算不对。。应该怎么算?
2个回答
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解:
令y-1=sint,则y=1+sint
y:0→2,则t:-π/2→π/2
∫[0:2]y(2-√(2y-y²)dy
=∫[0:2]2ydy-∫[0:2]y√[1-(y-1)²]dy
=∫[-π/2:π/2]2(1+sint)d(1+sint) -∫[-π/2:π/2](1+sint)costd(1+sint)
=∫[-π/2:π/2](2cost+2sintcost)dt -∫[-π/2:π/2](cos²t+sintcos²t)dt
=∫[-π/2:π/2]2costdt -∫[-π/2:π/2]cos²tdt
=4∫[0:π/2]costdt -∫[0:π/2](1+cos2t)dt
=4sint|[0:π/2] -(t+½sin2t)|[0:π/2]
=4(sinπ/2 -sin0) -[(π/2 +½sinπ)-(0+½sin0)]
=4- ½π
令y-1=sint,则y=1+sint
y:0→2,则t:-π/2→π/2
∫[0:2]y(2-√(2y-y²)dy
=∫[0:2]2ydy-∫[0:2]y√[1-(y-1)²]dy
=∫[-π/2:π/2]2(1+sint)d(1+sint) -∫[-π/2:π/2](1+sint)costd(1+sint)
=∫[-π/2:π/2](2cost+2sintcost)dt -∫[-π/2:π/2](cos²t+sintcos²t)dt
=∫[-π/2:π/2]2costdt -∫[-π/2:π/2]cos²tdt
=4∫[0:π/2]costdt -∫[0:π/2](1+cos2t)dt
=4sint|[0:π/2] -(t+½sin2t)|[0:π/2]
=4(sinπ/2 -sin0) -[(π/2 +½sinπ)-(0+½sin0)]
=4- ½π
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