解ï¼è¯¦ç»è¿ç¨æ¯ï¼âµåå¼=â«(0,Ï)dxâ«(0,sinx)(x²-y²)dyã
èï¼â«(0,sinx)(x²-y²)dy=[x²y-(1/3)y³]丨(y=0,sinx)=x²sinx-(1/3)sin³xã
â´åå¼=â«(0,Ï)[x²sinx-(1/3)sin³x]dx=â«(0,Ï)x²sinxdx-(1/3)â«(0,Ï)sin³xdxã
â«x²sinxdx=-â«x²d(cosx)=-x²cosx+2â«xcosxdx=-x²cosx+2xsinx-2â«sinxdx=-x²cosx+2xsinx +2cosx+Cã
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追çæ¥ä¸æ¥çè¿ç¨æ¯ï¼â´â«(0,Ï)x²sinxdx=[-x²cosx+2xsinx +2cosx]丨(x=0,Ï)=ϲ-4ã
â«(0,Ï)sin³xdx=-â«(0,Ï)(1-cos²x)d(cosx)=-[cosx-(1/3)cos³x]丨(x=0,Ï)=4/3ã
â´åå¼=ϲ-4-(1/3)4/3=ϲ-40/9ã
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