一道高数证明题，毫无头绪，求思路
设f(x)在[a,b]上二阶可导,且f”(x) <0,证明: ∫(a,b)f(x)dx≦(b-a)f[(a+b)/2]
[解决办法]
f(x)=f(c)+f '(c)(x-c)+f '(t)(x-c)^2 <= f(c)+f '(c)(x-c)
c=(a+b)/2 a <=t <=b
∫(a,b)f(x)dx <= ∫(a,b)(f(c)+f '(c)(x-c))dx=∫(a,b)f(c)dx+∫(a,b)f '(c)(x-c)dx
f(c)(b-a)+f '(c)((b-c)^2-(a-c)^2)=f(c)(b-a)=f((b+a)/2)(b-a)
