设f(x),g(x)在[a,b]上连续,且满足∫axf(t)dt≥∫axg(t)dt,x∈[a,b),∫abf(t)dt=∫abg(t)dt.证明:∫ab xf(x)dx≤∫ab xg(x)dx.

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问题 设f(x),g(x)在[a,b]上连续,且满足∫axf(t)dt≥∫axg(t)dt,x∈[a,b),∫abf(t)dt=∫abg(t)dt.证明:∫ab xf(x)dx≤∫ab xg(x)dx.

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答案当x∈[a,b)时, ∫axf(t)dt≥∫ax g(t)[*]∫ax[f(t)一g(t)]dt≥0, ∫ab f(t)dt=∫ab g(t)[*]∫ab[f(t)一g(t)]dt=0, ∫abxf(x)dx≤∫abxg(x)dx [*] ∫abx[f(x)一g(x)]dx≤0, 令G(x)=∫ax[f(t)一g(t)]dt,则G’(x)=f(x)一g(x),于是 ∫abx[f(x)一g(x)]dx=∫abxd(∫ax[f(x)一g(t)]dt) ∫abx[f(x)一 g(x) ]dx =∫ab xd (∫ax[f(t)一 g(t) dt) [*] x∫ax[f(t)一g(t)dt]|ab-∫ab[∫ax(f(t)-g(t))dt]dx =-∫ab[∫ax(f(t))-g(t)dt]dt≤0(因为G(x)一∫ax[f(t)-g(t)]dt≥0), 即 ∫ab x[f(x)一g(x)]dx≤0,即∫ab xf(x)dx≤∫ab xg(x)dx.

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