Sign of Function Matches Sign of Definite Integral

Theorem

Let $f$ be a real function continuous on some closed interval $\left [{a \,.\,.\, b} \right]$, where $a < b$.

Then:

If $\forall x \in \left [{a \,.\,.\, b} \right] : f\left({x}\right) \ge 0$ then $\displaystyle \int_a^b f\left({x}\right) \, \mathrm dx \ge 0$

If $\forall x \in \left [{a \,.\,.\, b} \right] : f\left({x}\right) > 0$ then $\displaystyle \int_a^b f\left({x}\right) \, \mathrm dx > 0$

If $\forall x \in \left [{a \,.\,.\, b} \right] : f\left({x}\right) \le 0$ then $\displaystyle \int_a^b f\left({x}\right) \, \mathrm dx \le 0$

If $\forall x \in \left [{a \,.\,.\, b} \right] : f\left({x}\right) < 0$ then $\displaystyle \int_a^b f\left({x}\right) \, \mathrm dx < 0$

Proof

From Continuous Function is Riemann Integrable, the definite integrals under discussion are guaranteed to exist.

Consider the case where $\forall x \in \left [{a \,.\,.\, b} \right] : f\left({x}\right) \ge 0$.

Define a constant mapping:

$\mathbf 0: \left [{a \,.\,.\, b} \right] \to \R$:

$\mathbf 0 \left({x}\right) = 0$

Then:

\(\displaystyle \mathbf 0\left({x}\right)\)

\(\le\)

\(\displaystyle f\left({x}\right)\)

for any $x \in \left [{a \,.\,.\, b} \right]$: recall $f\left({x}\right) \ge 0$

\(\displaystyle \implies \ \ \)

\(\displaystyle \int_a^b \mathbf 0 \left({x}\right) \, \mathrm dx\)

\(\le\)

\(\displaystyle \int_a^b f \left({x}\right) \, \mathrm dx\)

Relative Sizes of Definite Integrals

\(\displaystyle \implies \ \ \)

\(\displaystyle 0 \left({b-a}\right)\)

\(\le\)

\(\displaystyle \int_a^b f \left({x}\right) \, \mathrm dx\)

Integral of Constant

\(\displaystyle \implies \ \ \)

\(\displaystyle \int_a^b f \left({x}\right) \, \mathrm dx\)

\(\ge\)

\(\displaystyle 0\)

The proofs of the other cases are similar.

$\blacksquare$