设连续函数fx满足fx=x^2-∫(0∽2)fxdx,求∫(0∽2)fxdx= fx连续,且fx=x+2∫(0,1)ftdt,则fx= 要过...
\u8bbe\u8fde\u7eed\u51fd\u6570fx\u6ee1\u8db3fx=x\u22272-\u222b1-0 fxdx,\u6c42fxf(x)=x^2-\u222b(0->1) f(x)dx
let
f(x) =x^2+ C
x^2-\u222b(0->1) f(x)dx
=x^2-\u222b(0->1) (x^2+C)dx
=x^2 - [(1/3)x^3+ Cx]|(0->1)
=x^2 - (1/3+ C )
=>
C= -(1/3+C)
C= -1/6
f(x) = x^2 -1/6
\u4e24\u8fb9\u5bf9x\u6c42\u5bfc,\u5f97f'(x)+2f(x)=2x
\u518d\u4e24\u8fb9\u5bf9x\u6c42\u5bfc,\u5f97f''(x)+2f'(x)=2,\u4ee4t=f'(x),\u5219dt/dx=2-2t\u5373dt/(t-1)=-2dx
\u4e24\u8fb9\u79ef\u5206\u5f97ln[C(t-1)]=-2x,C\u4e3a\u5e38\u6570
\u5219f'(x)-1=[e^(-2x)]/C
\u79ef\u5206,\u5f97f(x)=D+x-[e^(-2x)]/(2C),C\u3001D\u4e3a\u5e38\u6570
\u800c\u9898\u4e2d\u5f0f\u5b50\u4ee5x=0\u4ee3\u5165,\u53ef\u5f97f(0)=0,\u6240\u4ee5D-1/(2C)=0
\u518d\u4ee5x=-1/2\u4ee3\u5165,\u5f97C=-1,\u90a3\u4e48D=-1/2-
\u5219f(x)=x-1/2+(e^(-2x))/2
\u6269\u5c55\u8d44\u6599\u4e0d\u5b9a\u79ef\u5206\u7684\u516c\u5f0f
1\u3001\u222b a dx = ax + C\uff0ca\u548cC\u90fd\u662f\u5e38\u6570
2\u3001\u222b x^a dx = [x^(a + 1)]/(a + 1) + C\uff0c\u5176\u4e2da\u4e3a\u5e38\u6570\u4e14 a \u2260 -1
3\u3001\u222b 1/x dx = ln|x| + C
4\u3001\u222b a^x dx = (1/lna)a^x + C\uff0c\u5176\u4e2da > 0 \u4e14 a \u2260 1
5\u3001\u222b e^x dx = e^x + C
6\u3001\u222b cosx dx = sinx + C
7\u3001\u222b sinx dx = - cosx + C
8\u3001\u222b cotx dx = ln|sinx| + C = - ln|cscx| + C
如图
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