Sign of Function Matches Sign of Definite Integral

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.

$\blacksquare$