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设f'(x)=arctan[(x-1)^2],f(0)=0,求∫(0,1)f(x)dx,其中0是下限,1是上限,π/8-1/4ln2,求详解设f'(x)=arctan[(x-1)^2],f(0)=0,求∫(0,1)f(x)dx,其中0是下限,1是上限,π/8-1/4ln2,求详解
题目详情
设f'(x)=arctan[(x-1)^2],f(0)=0,求∫(0,1)f(x)dx,其中0是下限,1是上限,π/8-1/4ln2,求详解
设f'(x)=arctan[(x-1)^2],f(0)=0,求∫(0,1)f(x)dx,其中0是下限,1是上限,
π/8-1/4ln2,求详解
设f'(x)=arctan[(x-1)^2],f(0)=0,求∫(0,1)f(x)dx,其中0是下限,1是上限,
π/8-1/4ln2,求详解
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答案和解析
原积分=积分（从0到1）f(x)d(x-1)=(x-1)f(x)|下限0上限1-积分（从0到1）(x-1)f'(x)dx
=-积分（从0到1）(x-1)arctan[(x-1)^2]dx=(变量替换1-x=t)积分（从0到1）t*arctan(t^2)dt=（变量替换t^2=x）0.5积分（从0到1）arctanxdx=0.5xarctanx|下限0上限1-0.5积分（从0到1）x/(1+x^2)dx=pi/8-0.25ln(1+x^2)|下限0上限1=pi/8-0.25ln2.
=-积分（从0到1）(x-1)arctan[(x-1)^2]dx=(变量替换1-x=t)积分（从0到1）t*arctan(t^2)dt=（变量替换t^2=x）0.5积分（从0到1）arctanxdx=0.5xarctanx|下限0上限1-0.5积分（从0到1）x/(1+x^2)dx=pi/8-0.25ln(1+x^2)|下限0上限1=pi/8-0.25ln2.
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