Relative Sizes of Definite Integrals

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$, where $a < b$.


If:

$\forall t \in \closedint a b: \map f t \le \map g t$

then:

$\displaystyle \int_a^b \map f t \rd t \le \int_a^b \map g t \rd t$


Similarly, if:

$\forall t \in \closedint a b: \map f t < \map g t$

then:

$\displaystyle \int_a^b \map f t \rd t < \int_a^b \map g t \rd t$


Proof

Suppose that $\forall t \in \closedint a b: \map f t \le \map g t$.

From the Fundamental Theorem of Calculus, $g - f$ has a primitive on $\closedint a b$.

Let $H$ be such a primitive.

Then:

$\forall t \in \closedint a b: D_t \map H t = \map g t - \map f t \ge 0$

By Derivative of Monotone Function, it follows that $H$ is increasing on $\closedint a b$.

Thus:

$\map H b \ge \map H a$

Hence:

$\displaystyle \int_a^b \map g t - \map f t \rd t = \map H b - \map H a \ge 0$

The proof for the second case is similar.

$\blacksquare$


Also see


Sources