Question

定积分求解

就是下面图的那道题，最好有步骤……

Answer 1

亲，木有图哟

下面这题给你参考！
计算定积分∫sinx二次方/（1+e^x）dx（上限pi/2下限-pi/2）

∫[-π/2,π/2] (sinx^2)dx/(1+e^x)
=(1/2)∫[-π/2,π/2] (1-cos2x)dx/(1+e^x)
=(1/2)∫[-π/2,π/2]d-e^(-x)/(1+e^(-x)+(-1/2)∫[-π/2, π/2] cos2xdx/(1+e^x)
=(1/2)( -ln(1+e^(-x)) )|[-π/2, π/2] +(-1/2)∫[-π/2, π/2]cos2xdx/(1+e^x)
=(1/2)ln(1+e^(-π/2))/(1+e^(π/2))] -(1/2)∫[-π/2,π/2] cos2xdx/(1+e^x)
=(1/2)ln|(1+e^(-π/2))/(1+e^(π/2))|
∫[-π/2,π/2] cos2xdx/(1+e^x)
=(-1/2)∫[-π/2, π/2]dsin2x/(1+e^x)
=-(1/2)sin2x/(1+e^x)| [-π/2, π/2] (-1/2)∫[-π/2,π/2]sin2xe^xdx/(1+e^x)^2
=(-1/2)∫[-π/2,0] sin2xe^xdx/(1+e^x)^2+(-1/2)∫[0,π/2] sin2xe^xdx/(1+e^x)^2
=(1/2)∫[0,-π/2]sin2xe^xdx/(1+e^x)^2 +(-1/2)∫[0,-π/2] sin2ue^(-u)du/(1+e^(-u))^2 （其中u=-x)
=(1/2)∫[0,-π/2]sin2xe^xdx/(1+e^x)^2+(-1/2)∫[0, -π/2]sin2ue^udu/(1+e^u)^2
=0

f(x)=sin2xe^x/(1+e^x)^2
f(-x)=sin(-2x)e^(-x)/(1+e^(-x))^2=-sin2xe^x/(1+e^x)^2=-f(x)
奇函数的对称积分为0 如：∫[-a,a] sinxdx=-cosa+cos(-a)=0

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Answer 2

Answer 3