设函数f(x)在[0,1]上连续,且满足f(x)=x^2-3x∫f(t)dt（上限为1，下限为0）,试求f(x) 诚求详细过程！ 设函数f(x)具有连续一阶导数，且满足f(x)=∫(上限是x...
\u8bbe\u51fd\u6570f(x)\u5728(0,1)\u4e0a\u8fde\u7eed,\u4e14\u6ee1\u8db3f(x)=x+2 \u222b(0,1)f(t)dt,\u6c42f(x)\u66f4\u7b80\u6d01\u7684\u8868\u8fbe\u5f0f\u4ee4a=\u222b(0,1)f(t)dt, \u5b83\u4e3a\u5e38\u6570
\u6545f(x)=x+2a
\u518d\u4ee3\u5165\u4e0a\u8ff0\u79ef\u5206\uff1a
a=\u222b(0,1)(t+2a)dt=(t^2/2+2at)|(0,1)=1/2+2a
\u89e3\u5f97\uff1aa=-1/2
\u6240\u4ee5f(x)=x-1
\u89e3\uff1af(x)=\u222b(\u4e0a\u9650\u662fx\u4e0b\u9650\u662f0)(x^2-t^2)f'(t)dt+x^2 \u6240\u4ee5f(0)=0,
\u53c8\u51fd\u6570f(x)\u5177\u6709\u8fde\u7eed\u4e00\u9636\u5bfc\u6570\uff0c\u5bf9\u4e0a\u5f0f\u4e24\u8fb9\u6c42\u5bfc\u5f97;
f'(x)=)=\u222b(\u4e0a\u9650\u662fx\u4e0b\u9650\u662f0)2xf'(t)dt+2x=2xf(x)+2x=2x(f(x)+1)
dy/(y+1)=2xdx \u89e3\u5f97f(x)=e^x^2-1.
\u6709\u95ee\u9898\u8bf7\u8ffd\u95ee \u6ee1\u610f\u8bf7\u53ca\u65f6\u91c7\u7eb3\u3002
设∫(上限1，下限0) f(t) dt= C，C为常数，
则f(x)= x^2 -3Cx
于是
∫(上限1，下限0) x^2 -3Cx dx
= (x^3)/3 -3C/2 *x^2，代入上下限1和0
=1/3 -3C/2
=C
解得C=2/15
所以f(x)=x^2 - 2x/5
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