计算二重积分∫∫e^y^2dσ,其中D：y=x及y=2x,y=1所围成的闭区域
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y=x及y=2x,y=1交点(1/2,1),(1,1)
则∫∫e^y^2dσ
=∫[0,1]∫[y/2,y]e^y^2dxdy
=∫[0,1]e^y^2∫[y/2,y]dxdy
=∫[0,1]e^y^2*y/2dy
=1/4∫[0,1]e^y^2dy^2
=1/4e^y^2[0,1]
=1/4(e-1)
======以下答案可供参考======
供参考答案1:
∫∫e^y^2dσ
D:{ y/2 { 0原式=∫(0,1)dy∫(y/2,y)e^(y²)dx
=1/2∫(0,1)ye^(y²)dy
=1/4∫(0,1)e^(y²)dy²
=1/4*(e^(y²))|(0,1)
=1/4(e-1)