Some Properties of Definite Integral

Prerequisite knowledge: Newton-Leibniz formula

Parity

set $f$在$[-a,a](a>0)$ , then

$${\int }_{-a}^{a}f(x)dx=\{\begin{array}{l}2{\int }_0^{a}f(x)dx,fis\; an\; even\; function\\ 0,fis\; an\; odd\end{array}$$

Proof: For ${\int }_{-a}^{0}f(x)dx$，令$x=-t$ , then

$${\int }_{-a}^{0}f(x)dx={\int }_a^{0}f(-t)d(-t)=-{\int }_a^{0}f(-t)dt={\int }_0^{a}f(-t)dt={\int }_0^{a}f(-x)dx$$

\qquad 所以原式$={\int }_0^{a}f(x)dx+{\int }_{-a}^{0}f\left(x\right)dx={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f(-x)dx$

$={\int }_0^a[f(x)+f(-x)]dx$

$=\{\begin{array}{l}2{\int }_0^{a}f(x)dx,fis\; an\; even\; function\\ 0,fis\; an\; odd\end{array}$

periodically

set $f$ is$T$ is a periodic continuous function, then for any real number$a$ , both have

$${\int }_a^{a+T}f(x)dx={\int }_0^{T}f\left(x\right)dx$$

证明：
$${\int }_a^{a+T}f\left(x\right)dx={\int }_a^{0}f\left(x\right)dx+{\int }_0^{T}f\left(x\right)dx+{\int }_T^{a+T}f\left(x\right)dx$$

\qquad Let $x=t+T$ , then

$${\int }_T^{a+T}f(x)dx={\int }_0^{a}f(t+T)dt={\int }_0^{a}f(t)dt$$

\qquad so

${\int }_a^{a+T}f\left(x\right)dx={\int }_a^{0}f\left(x\right)dx+{\int }_0^{T}f\left(x\right)dx+{\int }_T^{a+T}f\left(x\right)dx$

$={\int }_a^{0}f\left(x\right)dx+{\int }_0^{T}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$={\int }_0^{T}f\left(x\right)dx$