Let $f:[a,\infty) \rightarrow \mathbb{R}$ continuous and $f(t)\ge0$, prove $F(x)=\int_a^{x}f(t)dt$ is increasing

Question

Today i started working with improper integrals and after calculating some of these i tried to prove some of the porpositions or theroems involving improper integrals. Now i am trying to prove this: Let $f:[a,\infty] \rightarrow \mathbb{R}$ continuous and $f(t)\ge0$, we define $F(x)=\int_a^{x}f(t)dt$, prove $F(x)$ is increasing. Any hint on how can i prove this? Thanks.

Answer 1

For $x>y$, $F(x)-F(y)=\displaystyle\int_{y}^{x}f(t)dt\geq\int_{y}^{x}0dt=0$, so $F(x)\geq F(y)$.

Answer 2

A Hammer. Use the FTC to deduce that $$ F'(x)=f(x)\geq 0 $$ which implies that $F$ is increasing.

Answer 3

Let $$ x_1<x_2$$ then $$F(x_2)-F(x_1) = $$

$$ \int _a^ {x_2} f(t)dt -\int_{a}^ {x_1} f(t)dt \ge $$

$$ \int_{x_1}^ {x_2} f(t)dt \ge0 $$

Thus F(x) is increasing on $[a, \infty)$