Does an integral preserve strict inequality?

Question

Suppose that $f$ and $g$ are real valued continuous functions on $[0,1]$ such that $f(t)<g(t)$ for all $t\in[0,1]$. Is it true that $$\int_{0}^{1}f(t) \ dt<\int_{0}^{1}g(t) \ dt \ ?$$ I know that this result is true when we replace "$<$" by "$\leq$".

Answer 1

Yes, of course, because the integral of a positive continuous function is positive: $$\int_0^1g(x)dx-\int_0^1f(x)dx=\int_0^1(g(x)-f(x))dx>0.$$

Answer 2

lets suppose that: $$f(t)<g(t)\Rightarrow f(t)-g(t)<0$$ intuitively this would give: $$\int_0^1\left[f(t)-g(t)\right]dt<0$$ which is the same as: $$\int_0^1f(t)dt<\int_0^1g(t)dt$$ since integrals follow the property: $$\int\left[\alpha(x)+\beta(x)\right]dx=\int\alpha(x)dx+\int\beta(x)dx$$