Sign of Function Matches Sign of Definite Integral

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Theorem

Let $f$ be a real function continuous on some closed interval $\closedint a b$, where $a < b$.


Then:

If $\forall x \in \closedint a b: \map f x \ge 0$ then $\displaystyle \int_a^b \map f x \rd x \ge 0$
If $\forall x \in \closedint a b: \map f x > 0$ then $\displaystyle \int_a^b \map f x \rd x > 0$
If $\forall x \in \closedint a b: \map f x \le 0$ then $\displaystyle \int_a^b \map f x \rd x \le 0$
If $\forall x \in \closedint a b: \map f x < 0$ then $\displaystyle \int_a^b \map f x \rd x < 0$


Proof

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

Consider the case where $\forall x \in \closedint a b: \map f x \ge 0$.

Define a constant mapping:

$f_0: \closedint a b \to \R$:
$\map {f_0} x = 0$

Then:

\(\displaystyle \map {f_0} x\) \(\le\) \(\displaystyle \map f x\) for any $x \in \closedint a b$: recall $\map f x \ge 0$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int_a^b \map {f_0} x \rd x\) \(\le\) \(\displaystyle \int_a^b \map f x \rd x\) Relative Sizes of Definite Integrals
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0 \paren {b - a}\) \(\le\) \(\displaystyle \int_a^b \map f x \rd x\) Integral of Constant
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int_a^b \map f x \rd x\) \(\ge\) \(\displaystyle 0\)

The proofs of the other cases are similar.

$\blacksquare$