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及f(x)!
问题描述：
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及f(x)!
答
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第三四五步没看懂……第三步两边同时取定积分，下限0，上限1第四步化简定积分第五步合并AA=∫[0->1] f(t) dt = ∫[0->1] f(x) dx