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As defined, $f(x)$ is continuous in $[a,b]$, a closed interval, so its image is also a closed interval. Therefore, $\exists \alpha \in \mathbb{R}$ such that $f(x) \gt \alpha \gt 0 \forall x\in[a, b]$, so:

$$\int_a^bf(x)dx \gt \int_a^b\alpha dx = \alpha(b-a) \gt 0$$

As defined, $f(x)$ is continuous in $[a,b]$, a closed interval, so its image is also a closed interval. Therefore, $\exists \alpha \in \mathbb{R}$ such that $f(x) \gt \alpha \gt 0 \forall x\in[a, b]$, so:

$$\int_a^bf(x)dx \geq \int_a^b\alpha dx = \alpha(b-a) \gt 0$$

As defined, $f(x) \gt \alpha \gt 0$, so:

$$\int_a^bf(x)dx \gt \int_a^b\alpha dx = \alpha(b-a) \gt 0$$

As defined, $f(x)$ is continuous in $[a,b]$, a closed interval, so its image is also a closed interval. Therefore, $\exists \alpha \in \mathbb{R}$ such that $f(x) \gt \alpha \gt 0 \forall x\in[a, b]$, so:

$$\int_a^bf(x)dx \gt \int_a^b\alpha dx = \alpha(b-a) \gt 0$$

As defined, $f(x) \gt \alpha \gt 0$, so:

$$\int_a^bf(x)dx \geq \int_a^b\alpha dx = \alpha(b-a) \gt 0$$

As defined, $f(x) \gt \alpha \gt 0$, so:

$$\int_a^bf(x)dx \gt \int_a^b\alpha dx = \alpha(b-a) \gt 0$$