f(x)在[a,b]上连续可导,f'(x)≤0 若F(x)=1/x-a,定积分∫f(t)dt[a,x] 证明在(a,b)满足F'(x)≤0
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f(x)在[a,b]上连续可导,f'(x)≤0 若F(x)=1/x-a,定积分∫f(t)dt[a,x] 证明在(a,b)满足F'(x)≤0
如题,
答
F(x)=[∫[a->x]f(t)dt]/(x-a)=>F'(x)=[f(x)(x-a)-∫[a->x]f(t)dt]/(x-a)²∴只需证明f(x)(x-a)-∫[a->x]f(t)dt≤0而f'(x)≤0,∴t∈[a,x]时,有f(t)≥f(x),不等式两边对t从a积分到x,则有∫[a->x]f(t)dt≥∫[a->x]f...