Properties of Definite Integrals

Throughout, we assume $f$ and $g$ are arbitrary functions and $c$ is any constant:

$\int_a^b c \; dx = c(b-a)$. $\int_a^b [f(x) \pm g(x)] \,dx = \int_a^b f(x) dx \pm \int_a^b g(x) \,dx$ $\int_a^b cf(x) \,dx = c\int_a^b f(x) \,dx$ $\int_b^a f(x)\, dx = -\int_a^b f(x) \,dx$ $\int_a^a f(x) \,dx = 0$

If one function is bigger than another, then its definite integral from $a$ to $b$ (with $a \le b$) will also be bigger. In particular,

If $a \le b$ and $f(x) \geq 0$ for $a \leq x \leq b$, then $\int_a^b f(x) dx \geq 0$. If $a \le b$ and $f(x) \geq g(x)$ for $a \leq x \leq b$, then $\int_a^b f(x)\, dx \geq \int_a^b g(x)\, dx$. If $a \le b$ and $m \leq f(x) \leq M$ for $a \leq x \leq b$, then $$m(b-a) \leq \int_a^b f(x)\, dx \leq M(b-a)$$ (Why?)

In other words, we can compare $f$ to $0$, we can compare $f$ to $g$, and we can compare $f$ to $m$ and $M$.

These properties are easy to visualize if you think about area. If $f(x) \ge g(x)$, then the curve $y=f(x)$ lies above the curve $y=g(x)$, so there is more area under the $f$ curve than under the $g$ curve.

But these properties apply even when we are dealing with negative functions or applications that have nothing to do with area. For instance, if $f(x) \ge g(x)$ for all $x \in [a,b]$, then for each $i=1,\ldots,n$ $$f(x_i^*) \ge g(x_i^*) \quad \Longrightarrow \quad f(x_i^*)\,\Delta x \ge g(x_i^*)\,\Delta x.$$ Hence, $${\sum_{i=1}^n f(x_i^*)\, \Delta x} \ge {\sum_{i=1}^n g(x_i^*)\, \Delta x}.$$ By limit laws, $${\lim_{n \to \infty}\, \sum_{i=1}^n f(x_i^*)\, \Delta x} \ge {\lim_{n \to \infty}\,\sum_{i=1}^n g(x_i^*)\, \Delta x},$$ and by definition of integral $$\int_a^b f(x)\, dx \ge \int_a^b g(x)\, dx.$$