作业帮设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt 证明：在内有设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt证明：在(a,b)内有F'(x)
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![设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt 证明：在内有设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt证明：在(a,b)内有F'(x)](/uploads/image/z/2669361-33-1.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B4%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%28a%2Cb%29%E5%86%85%E5%8F%AF%E5%AF%BC%E4%B8%94f%27%28x%29%E2%89%A40%2CF%28X%29%3D1%5C%28x-a%29%C2%B7%E2%88%AB%EF%BC%9Ca%2Cx%EF%BC%9Ef%28t%29dt+%E8%AF%81%E6%98%8E%EF%BC%9A%E5%9C%A8%E5%86%85%E6%9C%89%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B4%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%28a%2Cb%29%E5%86%85%E5%8F%AF%E5%AF%BC%E4%B8%94f%27%28x%29%E2%89%A40%2CF%28X%29%3D1%5C%28x-a%29%C2%B7%E2%88%AB%EF%BC%9Ca%2Cx%EF%BC%9Ef%28t%29dt%E8%AF%81%E6%98%8E%EF%BC%9A%E5%9C%A8%28a%2Cb%29%E5%86%85%E6%9C%89F%27%28x%29)
设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt 证明：在内有设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt证明：在(a,b)内有F'(x)
设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt 证明：在内有
设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,
F(X)=1\(x-a)·∫＜a,x＞f(t)dt
证明：在(a,b)内有F'(x)≤0
＜a,x＞分别是积分的上下限,
设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt 证明：在内有设函数f(x）在区间[a,b]上连续,在(a,b)内可导且f'(x)≤0,F(X)=1\(x-a)·∫＜a,x＞f(t)dt证明：在(a,b)内有F'(x)
F'(x) = f(x)/(x-a)-∫ f(t)dt/(x-a)² = ((x-a)f(x)-∫ f(t)dt)/(x-a)².
在(a,b)上f'(x) ≤ 0,故f(x)单调减,f(x) ≤ f(t)对t∈(a,x)成立,于是∫ f(t)dt ≥ (x-a)f(x).
(x-a)f(x)-∫ f(t)dt ≤ 0,又(x-a)² > 0,故F'(x) = ((x-a)f(x)-∫ f(t)dt)/(x-a)² ≤ 0.