Sign of Function Matches Sign of Definite Integral

From ProofWiki
Jump to: navigation, search

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.

$\blacksquare$