Question

f(x)在[0,1]有连续的导数,f(0)=1,且∫∫f'(x+y)dxdy=∫∫f(t)dxdy,积分区域Dt={(x

Answer 1

∫∫f(t)dxdy=f(t)∫∫dxdy=t^2f(t)/2

∫∫f'(x+y)dxdy=∫(0,t)dv∫(0,t)f'(u)du=∫(0,t)(f(t)-1)dv=t(f(t)-1)

由t(f(t)-1)=t^2f(t)/2得：f(t)-1=tf(t)/2

f(t)=2/(2-t)

f(x)=2/(2-x) (0