For the first part assuming $\displaystyle 0 < t$,
$\displaystyle f''(t) \geq 0 \Rightarrow \int_0^t f''(t)\, dt \geq 0 \Rightarrow f'(t) \geq f'(0) > 0 \Rightarrow f'(t) > 0$
Rinse and repeat:
$\displaystyle f'(t) > 0 \Rightarrow \int_0^t f'(t)\, dt > 0 \Rightarrow f(t) > f(0)= 0 \Rightarrow f(t) > 0$
Now for $\displaystyle t_2 > t_1, f'(t) > 0 \Rightarrow \int_{t_1}^{t_2} f'(t)\, dt > 0 \Rightarrow f(t_2) > f(t_1)$
So f is monotonically increasing and yet f(0) = f(1) = 0. Thus f is identically 0 throughout [0,1]. Similarly for the next one, except
$\displaystyle \int_{t}^1 f''(t) \geq 0 \Rightarrow f'(t) \leq f'(1) \leq 0 \Rightarrow f'(t) < 0$
Then you can repeat what you did before
Hello,
An insignificant report...
I'm not sure it's correct to call the variable t in the integrand, just like the variable in the boundaries... I know it's understandable, but well, it's quite disturbing$\displaystyle
f''(t) \geq 0 \Rightarrow \int_0^t f''(t)\, dt \dots
$