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[Post New]23 Mar 2008 14:31:06 IST
    Subject: Re:prove the inequality........
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hsbhatt (5804)

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I think the inequalities have got mixed up.
 
f"(x)>=0 and f'(x)> 0 means
 
rac {f(b) - f(a)} {b-a} geq rac {f(x) - f(a)} {x-a}
 
Hence, f(x) leq rac{f(b) - f(a)} {b-a} (x-a) + f(a)
 
So, now  int_a^b f(x) dx leq int_a^b ( rac{f(b) - f(a)} {b-a} (x-a) + f(a)) dx
 
= (b-a) ( rac{f(a)+f(b)} {2})
 
By reversing the inequality for f"(x)<=0, you get the second result.
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