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f(2x+1)=xe^x,求定积分f(t)dt
人气:308 ℃ 时间:2019-12-18 23:09:18
解答
令: t=2x+1 ,则: dt=2dx , x=(t-1)/2
∫f(t)dt
= ∫f(2x+1) 2dx
=2∫xe^x dx
=2∫x de^x
=2 [xe^x -∫e^x dx] +C
=2 [xe^x -e^x ] +C
= 2*e^x*(x-1) +C
= 2*e^[(t-1)/2]*[t-3]/2 +C
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