f(x)=x²+∫(1,0)xf(t)dt+∫(2,0)f(t)dt求函数f(x)
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f(x)=x²+∫(1,0)xf(t)dt+∫(2,0)f(t)dt求函数f(x)
f(x)=x²+∫(1,0)xf(t)dt+∫(2,0)f(t)dt求函数f(x)
f(x)=x²+∫(1,0)xf(t)dt+∫(2,0)f(t)dt求函数f(x)
f(x)=x²+∫(1,0)xf(t)dt+∫(2,0)f(t)dt
令 ∫(1,0)f(t)dt = a ∫(2,0)f(t)dt = b
f(x)=x²+ax+b
对f(x)=x²+ax+b 两边积分 ∫(1,0)f(x)dx=∫(1,0)x²+ax+b dx
a= 1/3 + a/2 +b
对f(x)=x²+ax+b 两边积分 ∫(2,0)f(x)dx=∫(2,0)x²+ax+b dx
b= 8/3 + 2a+ 2b
解方程组 a/2 - b = 1/3
2a + b = - 8/3
解得 a= - 14/15 b= - 4/5
所以 f(x)= x² - 14x/15 - 4/5