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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)!
人气:109 ℃ 时间:2020-03-27 17:24:14
解答
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
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