2道高数题求解,并写出过程原因(1)(-1到1)∫dx/1+2^(1/x)=_____________(2)设(x→0)lim[a tanx+b(1-cosx)]/[c In(1-2x)+d(1-e^(-x^2))]=2 ,其中a^2+c^2不等于0,则必有 ( )A)b=4d B)b=-4d C)a=4c D)a=-4c
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![2道高数题求解,并写出过程原因(1)(-1到1)∫dx/1+2^(1/x)=_____________(2)设(x→0)lim[a tanx+b(1-cosx)]/[c In(1-2x)+d(1-e^(-x^2))]=2 ,其中a^2+c^2不等于0,则必有 ( )A)b=4d B)b=-4d C)a=4c D)a=-4c](/uploads/image/z/4048781-5-1.jpg?t=2%E9%81%93%E9%AB%98%E6%95%B0%E9%A2%98%E6%B1%82%E8%A7%A3%2C%E5%B9%B6%E5%86%99%E5%87%BA%E8%BF%87%E7%A8%8B%E5%8E%9F%E5%9B%A0%281%EF%BC%89%EF%BC%88-1%E5%88%B01%EF%BC%89%E2%88%ABdx%2F1%2B2%5E%281%2Fx%29%3D_____________%282%29%E8%AE%BE%28x%E2%86%920%EF%BC%89lim%5Ba+tanx%2Bb%281-cosx%29%5D%2F%5Bc+In%281-2x%29%2Bd%281-e%5E%28-x%5E2%29%29%5D%3D2+%2C%E5%85%B6%E4%B8%ADa%5E2%2Bc%5E2%E4%B8%8D%E7%AD%89%E4%BA%8E0%2C%E5%88%99%E5%BF%85%E6%9C%89+%EF%BC%88+%EF%BC%89A%EF%BC%89b%3D4d+B%EF%BC%89b%3D-4d+C%EF%BC%89a%3D4c+D%EF%BC%89a%3D-4c)
2道高数题求解,并写出过程原因(1)(-1到1)∫dx/1+2^(1/x)=_____________(2)设(x→0)lim[a tanx+b(1-cosx)]/[c In(1-2x)+d(1-e^(-x^2))]=2 ,其中a^2+c^2不等于0,则必有 ( )A)b=4d B)b=-4d C)a=4c D)a=-4c
2道高数题求解,并写出过程原因
(1)(-1到1)∫dx/1+2^(1/x)=_____________
(2)设(x→0)lim[a tanx+b(1-cosx)]/[c In(1-2x)+d(1-e^(-x^2))]=2 ,其中a^2+c^2不等于0,则必有 ( )
A)b=4d B)b=-4d C)a=4c D)a=-4c
2道高数题求解,并写出过程原因(1)(-1到1)∫dx/1+2^(1/x)=_____________(2)设(x→0)lim[a tanx+b(1-cosx)]/[c In(1-2x)+d(1-e^(-x^2))]=2 ,其中a^2+c^2不等于0,则必有 ( )A)b=4d B)b=-4d C)a=4c D)a=-4c
(1)原积分=∫(-1到0)1/[1+2^(1/x)]×dx+∫(0到1)1/[1+2^(1/x)]×dx
令前一个积分的x=﹣u,则原积分=∫(1到0)1/[1+2^(-1/u)]×d(-u)+∫(0到1)1/[1+2^(1/x)]×dx
=∫(0到1)1/[1+2^(-1/u)]×du+∫(0到1)1/[1+2^(1/x)]×dx
=∫(0到1)2^(1/u)/[2^(1/u)+1]×du+∫(0到1)1/[1+2^(1/x)]×dx
再令后一个积分的x=u,则原积分=∫(0到1)2^(1/u)/[2^(1/u)+1]×du+∫(0到1)1/[1+2^(1/u)]×du
=∫(0到1)[1+2^(1/u)]/[1+2^(1/u)]×du
=∫(0到1)du
=1.
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