证明题f(u,v)在区域D=上连续,证明∫(π/2)(0)f(sinx,cosx)dx=∫(π/2)(0)f(cosx,
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证明题
f(u,v)在区域D=上连续,证明∫(π/2)(0)f(sinx,cosx)dx=∫(π/2)(0)f(cosx,sinx)dx
就是定积分的上限是π/2,下限是0
顺便问下曲线y=1/x+ln(1+e^x)分别有哪几条渐近线
f(u,v)在区域D=上连续,证明∫(π/2)(0)f(sinx,cosx)dx=∫(π/2)(0)f(cosx,sinx)dx
就是定积分的上限是π/2,下限是0
顺便问下曲线y=1/x+ln(1+e^x)分别有哪几条渐近线
![证明题f(u,v)在区域D=上连续,证明∫(π/2)(0)f(sinx,cosx)dx=∫(π/2)(0)f(cosx,](/uploads/image/z/4225937-41-7.jpg?t=%E8%AF%81%E6%98%8E%E9%A2%98f%28u%2Cv%29%E5%9C%A8%E5%8C%BA%E5%9F%9FD%3D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E8%AF%81%E6%98%8E%E2%88%AB%28%CF%80%2F2%29%280%29f%28sinx%2Ccosx%29dx%3D%E2%88%AB%28%CF%80%2F2%29%280%29f%28cosx%2C)
证明:由于sinx,cosx是连续函数,而由已知f(u,v)在区域D=上连续,所以复合函数f(sinx,cosx)和f(cosx,sinx)是在0≤x≤π/2是连续的,因此在0≤x≤π/2上f(sinx,cosx)和f(cosx,sinx)积分都存在.做积分变换y=π/2-x有
∫(π/2)(0)f(sinx,cosx)dx=-∫(0)(π/2)f(sin(π/2-y),cos(π/2-y))dy=∫(π/2)(0)f(cosy,siny)dy=∫(π/2)(0)f(cosx,sinx)dx
证毕.
另外,曲线y=1/x+ln(1+e^x)有两条渐近线x=0和y=0.
∫(π/2)(0)f(sinx,cosx)dx=-∫(0)(π/2)f(sin(π/2-y),cos(π/2-y))dy=∫(π/2)(0)f(cosy,siny)dy=∫(π/2)(0)f(cosx,sinx)dx
证毕.
另外,曲线y=1/x+ln(1+e^x)有两条渐近线x=0和y=0.
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