设区域D由曲线y=sinx，x=±π2，y=1围成，则∫∫D(x5y−1)dxdy=（）A．πB．2C．-2D．-π

Question

设区域D由曲线y=sinx，x=±(π)/(2)，y=1围成，则

∫∫

D

(x5y−1)dxdy=（　　）

A．π
B．2
C．-2
D．-π

Answer 1

∵D={（X，Y）|−(π)/(2)≤x≤(π)/(2)，sinx≤y≤1}
∴

∫∫

D

(x5y−1)dxdy=

∫	(π)/(2) −(π)/(2)

dx

∫

1 sinx

(x5y−1)dy
=

∫	(π)/(2) −(π)/(2)

[(1)/(2)x5y2−y

]

1 y＝sinx

dx
=

∫	(π)/(2) −(π)/(2)

((1)/(2)x5−(1)/(2)x5sin2x+sinx−1)dx
=(1)/(2)

∫	(π)/(2) −(π)/(2)

x5dx−(1)/(2)

∫	(π)/(2) −(π)/(2)

x5sin2xdx+

∫	(π)/(2) −(π)/(2)

sinxdx−

∫	(π)/(2) −(π)/(2)

dx
=−

∫	(π)/(2) −(π)/(2)

dx
=-π
故选：D．