依题意可得, Jo f(x)dx=e-1-f()f(0)=0 方程两边求导并化简得, e ∫(x)+∫(x)=e y7f(r f(x)=e(edx+C) ex(edx+C) 中c +ce x 由f(0)=0得C 2’∴∫(x) x∠已 e 2o x y = −1 x y e y = f (x) M N P1 P2 依题意可得, ( ) 1 ( ) 0 f x dx e f x x x  = − − 且f (0) = 0 方程两边求导并化简得, x f (x) + f (x) = e f (x) e ( e e dx C) dx x dx +    =  − ( ) 2 e e dx C x x =  + − x x e Ce− = + 2 1 , 2 1 由f (0) = 0得C = − ( ). 2 1 ( ) x x f x e e −  = −