设f(t)=∫e^(-x^2)dx,求∫tf(t)dt=?？
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let dF(x) =e^(-x^2) dx
f(t) =∫(1->t^2) e^(-x^2) dx
= F(t^2)- F(1)
f'(t) = 2tF'(t^2)
=2te^(-t^4)
∫(0->1) tf(t)dt
=(1/2)∫(0->1) f(t)dt^2
=(1/2)t^2f(t) |(0->1) - (1/2)∫(0->1) t^2 f'(t) dt
=-(1/2)∫(0->1) t^2 f'(t) dt
=-∫(0->1) t^3.e^(-t^4) dt
=(1/4) e^(-t^4) |(0->1)
=(1/4) [ e^(-1) -1],4,
f(t) =∫(1->t^2) e^(-x^2) dx
= F(t^2)- F(1)
f'(t) = 2tF'(t^2)
=2te^(-t^4)
∫(0->1) tf(t)dt
=(1/2)∫(0->1) f(t)dt^2
=(1/2)t^2f(t) |(0->1) - (1/2)∫(0->1) t^2 f'(t) dt
=-(1/2)∫(0->1) t^2 f'(t) dt
=-∫(0->1) t^3.e^(-t^4) dt
=(1/4) e^(-t^4) |(0->1)
=(1/4) [ e^(-1) -1],4,
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