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:
| \(\displaystyle \mathbf 0\left({x}\right)\) | \(\le\) | \(\displaystyle f\left({x}\right)\) | $\quad$ for any $x \in \left [{a \,.\,.\, b} \right]$: recall $f\left({x}\right) \ge 0$ | $\quad$ | |||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle \int_a^b \mathbf 0 \left({x}\right) \, \mathrm dx\) | \(\le\) | \(\displaystyle \int_a^b f \left({x}\right) \, \mathrm dx\) | $\quad$ Relative Sizes of Definite Integrals | $\quad$ | ||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle 0 \left({b-a}\right)\) | \(\le\) | \(\displaystyle \int_a^b f \left({x}\right) \, \mathrm dx\) | $\quad$ Integral of Constant | $\quad$ | ||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle \int_a^b f \left({x}\right) \, \mathrm dx\) | \(\ge\) | \(\displaystyle 0\) | $\quad$ | $\quad$ |
The proofs of the other cases are similar.
$\blacksquare$
