# Relative Sizes of Definite Integrals

## Theorem

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

If:

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

then:

$\displaystyle \int_a^b f \left({t}\right) \rd t \le \int_a^b g \left({t}\right) \rd t$

Similarly, if:

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

then:

$\displaystyle \int_a^b f \left({t}\right) \rd t < \int_a^b g \left({t}\right) \rd t$

## Proof

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

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:

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

The proof for the second case is similar.

$\blacksquare$