设f(t)在[1,+∞)上连续,f(1)=0,积分上限x^3 下限1 f(t)dt=lnx 则f(e)=?求详解
问题描述:
设f(t)在[1,+∞)上连续,f(1)=0,积分上限x^3 下限1 f(t)dt=lnx 则f(e)=?
求详解
答
∫(1->x^3) f(t)dt =lnx
3x^2f(x^3) = 1/x
f(x^3)= (1/3)(1/x^3)
=>
f(x) = (1/3)(1/x)
f(e) =(1/3)(1/e)