Relative Sizes of Definite Integrals
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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:
- $\ds \int_a^b \map f t \rd t \le \int_a^b \map g t \rd t$
Similarly, if:
- $\forall t \in \closedint a b: \map f t < \map g t$
then:
- $\ds \int_a^b \map f t \rd t < \int_a^b \map g t \rd t$
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:
- $\ds \int_a^b \map g t - \map f t \rd t = \map H b - \map H a \ge 0$
The proof for the second case is similar.
$\blacksquare$
Also see
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 13.23$