Question

定积分问题在线等求详解！！

解答过程要有(>_<)

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Answer 1

1、原式=(-1/2)*∫(0,√2a) d(3a^2-x^2)*(3a^2-x^2)^(-1/2)
=(-1/2)*2*√(3a^2-x^2)|(0,√2a)
=(√3-1)a
2、因为被积函数是奇函数，且积分区域[-π,π]是关于原点对称的，所以积分值为0
3、当x>=1时，f(x-1)=1/x；当x<1时，f(x-1)=1/[e^(x-1)+1]
原式=∫(0,1)f(x-1)dx+∫(1,2)f(x-1)dx
=∫(0,1)dx/[e^(x-1)+1]+∫(1,2)dx/x
∫(1,2)dx/x=ln|x||(1,2)=ln2
令e^(x-1)+1=t x=ln(t-1)+1 dx=dt/(t-1)
∫(0,1)dx/[e^(x-1)+1]=∫(1/e+1,2)dt/(t-1)t
=∫(1/e+1,2) [1/(t-1)-1/t] dt
=[ln|t-1|-ln|t|]|(1/e+1,2)
=-ln2-[-1-ln(1/e+1)]
=1+ln(1/2e+1/2)
=ln[(e+1)/2]
所以原式=ln[(e+1)/2]+ln2=ln(e+1)