if f(x) is continous on interval [0 2]
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
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.