Question

这个变限积分对t求导怎么求？

谢谢。

Answer 1

令 u = xt，则 x = u/t, dx = du/t
F(t) = ∫<0, t>sin[(xt)^2]dx = ∫<0, t^2>sin[u^2]du/t
= (1/t)∫<0, t^2>sin[u^2]du
F'(t) = (-1/t^2)∫<0, t^2>sin[u^2]du + (1/t)2tsin(t^4)
= (-1/t^2)∫<0, t^2>sin[u^2]du + 2sin(t^4)

Answer 2

let
u=xt
du= tdx
x=0, u=0
x=t, u=t^2
y
=∫(0->t) sin(xt)^2 dx
=∫(0->t^2) sin(u^2) (du/t)
=(1/t)∫(0->t^2) sin(u^2) du
dy/dt
=(1/t)sin(t^4).(2t) -(1/t^2)∫(0->t^2) sin(u^2) du
=2sin(t^4) -(1/t^2)∫(0->t^2) sin(u^2) du

Answer 3

把式子里x换成t，乘t的导数1