解答题
24.(1)求
∫0x2xf(x-t)dt．
(2)设
，求df(x)｜x=1．
(3)设F(x)=∫0xdy∫0y2
∫0x2xf(x-t)dt．(2)设
，求df(x)｜x=1．(3)设F(x)=∫0xdy∫0y2
【正确答案】(1)由∫0x2xf(x-t)dt=x∫0x2f(x-t)dt
x∫0x-x2f(u)(-du)=x∫x-x2xf(u)du得
∫0x2xf(x-t)dt=∫x-x2xf(u)du+x[f(x)-(1-2x)f(x-x2)]．
(2)由f(x)=
=xex得
f'(x)=(x+1)ex，从而f'(1)=2e，故df(x)｜x=1=2edx．
(3)F'(x)=∫0x2
dt，F''(x)=
x∫0x-x2f(u)(-du)=x∫x-x2xf(u)du得∫0x2xf(x-t)dt=∫x-x2xf(u)du+x[f(x)-(1-2x)f(x-x2)]．(2)由f(x)=
=xex得f'(x)=(x+1)ex，从而f'(1)=2e，故df(x)｜x=1=2edx．
(3)F'(x)=∫0x2
dt，F''(x)=【答案解析】