Relative Sizes of Definite Integrals

Theorem
Let $f$ and $g$ be real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.

Suppose that $\forall t \in \left[{a \,.\,.\, b}\right]: f \left({t}\right) \le g \left({t}\right)$.

Then:
 * $\displaystyle \int_a^b f \left({t}\right) \ \mathrm d t \le \int_a^b g \left({t}\right) \ \mathrm d t$

Proof
From the Fundamental Theorem of Calculus, $g - f$ has a primitive on $\left[{a \,.\,.\, b}\right]$.

Let $H$ be such a primitive.

Then $\forall t \in \left[{a \,.\,.\, b}\right]: D_t H \left({t}\right) = g \left({t}\right) - f \left({t}\right) \ge 0$.

By Derivative of Monotone Function, it follows that $H$ is increasing on $\left[{a \,.\,.\, b}\right]$.

Thus $H \left({b}\right) \ge H \left({a}\right)$.

Hence $\displaystyle \int_a^b g \left({t}\right) - f \left({t}\right) \ \mathrm d t = H \left({b}\right) - H \left({a}\right) \ge 0$.

The result follows.