Question

大一高数题求解

大一高数题求解如图，求大神解答

Answer 1

x∫(0->1) f(tx)dt =xf(x) + x^2.sinx, f(0)=0
To find : f(x)
Solution :
let
u=tx
du = x dt
t=0, u=0
t=1, u=x
∫(0->1) f(tx)dt =∫(0->x) (1/x) f(u) du
x∫(0->1) f(tx)dt =∫(0->x) f(u) du
x∫(0->1) f(tx)dt =xf(x) + x^2.sinx
∫(0->x) f(u) du =xf(x) + x^2.sinx
两边取导
f(x) = f(x) + xf'(x) + x^2.cosx +2x.sinx
xf'(x) + x^2.cosx +2x.sinx=0
f'(x) = -(xcosx +2sinx)
f(x)
= ∫-(xcosx +2sinx) dx
= 2cosx - ∫xdsinx
=2cosx - xsinx +∫sinx dx
=2cosx - xsinx - cosx + C
=cosx -xsinx +C
f(0)=0
cos0 - 0 + C=0
C=-1
f(x) =cosx -xsinx -1