Question

若f(x) 连续,∫[0,1]xf(t)dt=f(x)+xe^x,求f(x)

Answer 1

∫(0->1)xf(t)dt=f(x)+xe^x
f(x) =-xe^x + ∫(0->1)xf(t)dt (1)
∫(0->1) f(x) dx = ∫(0->1) [-xe^x + ∫(0->1)xf(t)dt ] dx
=∫(0->1) (-xe^x) dx + [x^2/2](0->1) . ∫(0->1)f(t)dt
=∫(0->1) -x d(e^x) + (1/2)∫(0->1)f(t)dt
(1/2)∫(0->1) f(x) dx =[-xe^x](0->1) +∫(0->1) e^x dx
= -e + (e-1)
= -1
∫(0->1) f(x) dx = -2
from (1)
f(x) =-xe^x + ∫(0->1)xf(t)dt
=-xe^x -2x