if f(x) is continous on interval [0 2] then:
the integral from 0 to 2 of f(x) is no doubt positive just because of the fact that, the blotch on the negative portion of the y axis, does not substantiate the seen positive area under the curve from [0 2].
 
the integral from [0 2] of f '(x) is I believe negative.  This is because the integral of    f'(x) from [0 2] is the same as f(2)-f(0).  since f(2) is smaller than f(0), the integral is negative.
 
To find the integral from [0 2] of f''(x) I kind of used the same bit of logic as the above problem.  it is the same as f '(2)-f '(0). Since the slope @2 is negative but steeper than the apparent slope @ 0 then im going to have to say the integral is also negative.  s2<0<s1 no matter what   s2-s1 will be negative.
 
Jesse Poulin