Hmm, is this correct?

f'(c)(b-a)= ∫[a,b] f(t) dt

Let f(t) = 1/(1+t^2)

b=x and a=0

f'(c)(x-0)= ∫[0,x] 1/(1+t^2) dt = arctan(x)

For 0<c<x,

arctan(x) = x/(1+c^2)

since c<x and since numbers are positive,

x/(1+c^2) > x/(1+x^2) and thus arctan(x)>x/(1+x^2)