求变限积分∫（t^2-1）e^tdx的最值,积分上限是x^2.下限是0
2个回答

F(x)=∫[0,x^2]（t^2-1）e^tdt

= ∫（t^2-1）de^t [0,x^2]

=(t^2-1)e^t-∫2t e^t dt [0,x^2]

=(t^2-1)e^t-∫2t de^t [0,x^2]

=(t^2-1)e^t-2te^t+2e^t+C [0,x^2]

=e^t (t^2+1-2t) +C [0,x^2]

=e^(x^2) (x^4-2x^2)

F'(x)=e^(x^2) (4x^3-4x+2x^5-4x^3)=e^(x^2) (2x^5-4x)=0

x=0,+/-2^(1/4),x^2=0,2^0.5

F(0)=0,F(+/-2^0.25)=e^(2^0.5) *(2-2*2^0.5)

Fmax=e^(2^0.5) *(2-2*2^0.5)

Fmin=0

相关问题