设A = ∫[0->1] f(t) dt
f(x) = x²A + 3
∫[0->1] f(x) dx = A∫[0->1] x² dx + 3∫[0->1] dx
A = A(1/3) + 3
A = 9/2 = ∫[0->1] f(x) dx
f(x) = (9/2)x² + 3
设f(x)在区间[0,1]上连续,且满足f(x)=x²∫(0,1)f(t)dt+3,求∫(0,1)f(x)dx
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