Comparison of Definite Integrals

Let \( f(x) \) and \( g(x) \) be two functions that are Riemann integrable over the interval [a, b]. If \( f(x) \le g(x) \) for all \( x \in [a, b] \), then $$ \int_a^b f(x)\, dx \le \int_a^b g(x)\, dx $$

Proof

Assume \( f(x) \) and \( g(x) \) are integrable over [a, b], with \( a < b \).

Since \( f(x) \le g(x) \) on the entire interval, it follows that:

$$ \int_a^b f(x)\, dx \le \int_a^b g(x)\, dx $$

This inequality reflects the fact that the upper Darboux sums of \( f \) are less than or equal to those of \( g \):

$$ S(x, f) \le S(x, g) $$

We now examine the difference between the two integrals:

$$ \int_a^b g(x)\, dx - \int_a^b f(x)\, dx $$

By the linearity of integration, this is equivalent to:

$$ \int_a^b [g(x) - f(x)]\, dx $$

Since \( g(x) - f(x) \ge 0 \) over [a, b], its integral must also be non-negative. In terms of Darboux sums, the inequality \( S(x, g) - S(x, f) \ge 0 \) holds for every \( x \in [a, b] \), so in the limit as \( \Delta x \to 0 \):

$$ \lim_{\Delta x \to 0} [S(x, g) - S(x, f)] \cdot \Delta x \ge 0 $$

Therefore, we conclude:

$$ \int_a^b g(x)\, dx - \int_a^b f(x)\, dx \ge 0 $$

which implies:

$$ \int_a^b f(x)\, dx \le \int_a^b g(x)\, dx $$

Note: If a function is identically zero over [a, b], with \( a < b \), then $$ f(x) = 0 $$ and its definite integral is zero as well: $$ \int_a^b f(x)\, dx = 0 $$ If \( f(x) \ge 0 \) on [a, b], then its integral must be non-negative: $$ \int_a^b f(x)\, dx \ge 0 $$ For any function \( f(x) \), we always have the pointwise inequality: $$ -|f(x)| \le f(x) \le |f(x)| $$ Integrating both sides gives: $$ \int_a^b -|f(x)|\, dx \le \int_a^b f(x)\, dx \le \int_a^b |f(x)|\, dx $$

An Illustrative Example

Consider the functions:

$$ f(x) = x \qquad g(x) = 2x $$

Over the interval [1, 5], we clearly have:

$$ f(x) < g(x) $$

Let’s compute the definite integrals of both functions over [1, 5]:

$$ \int_1^5 f(x)\, dx = \int_1^5 x\, dx = \frac{1}{2} (5^2 - 1^2) = \frac{24}{2} = 12 $$

$$ \int_1^5 g(x)\, dx = \int_1^5 2x\, dx = 2 \cdot \frac{1}{2} (5^2 - 1^2) = 2 \cdot 12 = 24 $$

Thus, the inequality between the functions is preserved under integration:

$$ \int_1^5 f(x)\, dx < \int_1^5 g(x)\, dx $$

since

$$ 12 < 24 $$

And so on.