∫e^x[(1-cosy)dx-(y-siny)dy],其中c为区域 0≤x≤π,0≤y≤sinx的边界曲线取正向.求曲线积分
P(x,y)=e^x(1-cosy) -对y求偏导数=e^xsiny
Q(x,y)=e^x(siny-y) -->对x求偏导数=e^xsiny-ye^x
I=∫∫(e^xsiny-ye^x-e^xsiny)dxdy
=-∫∫(ye^x)dydy
=-[ ∫[0,π]e^xdx * ∫[0,sinx]ydy ]
=-(1/2)∫[0,π](sin^2x)e^xdx
因为sin^2x =1/2(cos2x-1)
I=-(1/2)∫[0,π](1/2(cos2x-1))e^xdx
=1/4[ ∫[0,π]e^cos2x dx - ∫[0,π]e^x dx ]
又∫e^cos2x dx =(e^xcos2x+2e^xsin2x)/5
I=1/4(1-e^π)
正确答案是1/5(1-e^π).
请指出我哪里计算错了.给出正确步骤
人气:403 ℃ 时间:2020-01-29 18:50:28
解答
在你写的“因为”以上的部分都是正确的sin²x=1/2(1-cos2x),你的这个公式写错了,最后的代入数字也算错了.I=-(1/4)∫[0,π] (1-cos2x)e^xdx=-(1/4)∫[0,π] e^xdx+(1/4)∫[0,π] e^xcos2xdx=-(1/4)e^x+(1/4)(1/5)...
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