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

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

选项

答案当x∈[a,b)时, ∫axf(t)dt≥∫axg(t)dt[*]∫ax[f(t)一g(t)]dt≥0, ∫abf(t)dt=∫abg(t)dt[*]∫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(t)一g(t)]dt)[*] x∫ax[f(t)-g(t)]dt|ab-∫ab{∫ax[f(t)-g(t)]dt}dx =一∫ab{∫ax[f(t)一g(t)]dt}dx≤0(G(x)=∫ax[f(t)一g(t)]dt≥0), 即∫ab[f(x)-g(x)]dx≤0,即∫abxf(x)dx≤∫abxg(x)dx.

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