Question

设f(x)有一阶导数且满足∫0到1f(tx)dt=f(x)+xsinx,则f(x)=

Answer 1

简单分析一下，答案如图所示

Answer 2

令 tx = u, 则 t = u/x, t = 0 时， u = 0； t = 1 时，u = 下。 dt = du/x
∫<0，1> f(tx)dt = f(x)+xsinx 化为
(1/x)∫<0，x> f(u)du = f(x) + xsinx，
∫<0，x> f(u)du = xf(x) + x^2 sinx
两边对 x 求导，得
f(z) = f(x) + xf'(x) + 2xsinx - x^2cosx
x ≠ 0, 得 f'(x) = xcosx - 2sinx
f(x) = ∫xcosxdx - 2∫sinxdx = ∫xdsinx + 2cosx
= xsinx - ∫sinxdx + 2cosx = xsinx + 3cosx + C