Properties of Definite Integrals

# Properties of Definite Integrals

We will now look at some important properties of definite integrals. We will not prove all parts to Theorem 1 below, but the reader is advised to browse through them.

Theorem 1: If $\int_a^b f(x) \: dx$ and $\int_a^b g(x) \: dx$ exist, and $c$ is a constant, then the following hold:a) $\int_{a}^{b}[ f(x) + g(x) ] \: dx = \int_{a}^{b}f(x) \: dx + \int_{a}^{b}g(x) \: dx$ (Addition Property).b) $\int_{a}^{b}[ f(x) - g(x) ] \: dx = \int_{a}^{b}f(x) \: dx - \int_{a}^{b}g(x) \: dx$ (Subtraction Property).c) $\int_{a}^{b} c f(x) \: dx = c \int_{a}^{b}f(x) \: dx$ (Multiple Property).d) If $f(x) = k$ where $k \in \mathbb{R}$, then $\int_{a}^{b} k \,dx = k(b - a)$ (Definite Integral of a Constant Property).e) $\int_{a}^{b}f(x) \: dx + \int_{b}^{c}f(x) \: dx = \int_{a}^{c} f(x) \: dx$. |

**Proof of d)**Let $f(x) = k$ where $k$ is a real number. Then:

\begin{align} \int_{a}^{b} c \,dx = c(b - a) \\ = \left. cx \, \right |^{b}_{a} \\ = cb - ca \\ = c(b - a) \end{align}

Theorem 2: If $n ≤ f(x) ≤ N$ for all $a ≤ x ≤ b$ then $n(b - a) ≤ \int_{a}^{b}f(x) \: dx ≤ N(b - a)$. |

**Proof:**Since $n ≤ f(x) ≤ N$ for all $x \in [a, b]$ then we have that $f$ is bounded on $[a, b]$. By the completeness property of the real numbers, there exists a least element and a greatest element, call them $n'$ and $N'$ such that $n' ≤ f(x) ≤ N'$ for all $x \in [a, b]$. Thus $n ≤ n' ≤ f(x) ≤ N' ≤ N$.

- Now by Theorem 1 Part D above, $\int_a^b n \: dx = n(b - a)$ and $\int_a^b n' \: dx = n'(b - a)$. Since $n ≤ n'$, this implies that $n(b - a) ≤ n'(b - a) = \int_a^b n' \: dx ≤ \int_a^b f(x) \: dx$. So half of the inequality above.

- Similarly, we also have that $\int_a^b N' \: dx = N'(b - a)$ and $\int_a^b N \: dx = N(b - a)$. Since $N' ≤ N$, this implies that $\int_a^b f(x) ≤ \int_a^b N' \: dx = N'(b-a) ≤ N(b-a)$.

- Putting both inequalities together and we have that $n(b-a) ≤ \int_a^b f(x) \: dx ≤ N(b-a)$. $\blacksquare$