定积分：设f(x)在区间[a,b]上有连续函数,且f(a)=f(b)=0, ∫ (b,a)f^2(x)dx=1

定积分：设f(x)在区间[a,b]上有连续函数,且f(a)=f(b)=0, ∫ (b,a)f^2(x)dx=1，证明：∫(b,a)xf(x)f'(x)dx=-1/2.

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