Here is my approach to it, I hope this helps:

So here you are starting off with the equation,

\(\displaystyle \int_{0}^{t_{1}}f(t)\,dt=\int_{0}^{t_{2}}g(t)\,dt \).

If \(\displaystyle g(t) ≥ f(t) \) for all \(\displaystyle t \), then:

\(\displaystyle g(t)=f(t)+m(t) \)

where \(\displaystyle m(t) \) is an arbitrary function which makes the two equations above true.

In case that \(\displaystyle g(t)=f(t) \), \(\displaystyle m(t)=0 \).

So now we can try substituting that into your original equation and we get,

\(\displaystyle \int_{0}^{t_{1}}f(t)\,dt=\int_{0}^{t_{2}}f(t)\,dt+\int_{0}^{t_{2}}m(t)\, dt \).

Move over the \(\displaystyle \int_{0}^{t_{2}}f(t)\, dt \) to the left side and what's left is:

\(\displaystyle \int_{0}^{t_{1}}f(t)\,dt\, -\int_{0}^{t_{2}}f(t)\,dt\,=\int_{0}^{t_{2}}m(t)\, dt \).

If \(\displaystyle f(t) \) and \(\displaystyle g(t) \) are non-negative functions and \(\displaystyle g(t)≥f(t) \) for all \(\displaystyle t \), \(\displaystyle m(t) \) has to be a non-negative function as well, and the previous equation is only true when \(\displaystyle t_{1} ≥ t_{2} \). If \(\displaystyle t_{2} > t_{1} \) (which would therefore make \(\displaystyle f(t)≥g(t) \)), we would get \(\displaystyle -\int_{0}^{t_{2}}m(t)\, dt \). However, \(\displaystyle m(x) \) is not a negative function in our case, so its integral cannot be negative if all previous derivations hold true.

And of course, if \(\displaystyle t_{1} = t_{2} \), then:

\(\displaystyle \int_{0}^{t_{1}}f(t)\,dt\, -\int_{0}^{t_{2}}f(t)\,dt\,=0 \).

Hope that helped, and feel free to ask any questions!