## Even >[!abstract] Thereom >A function $f(x)$ is [[Even]] if $f(-x) = f(x)$ for all $x$ in the [[Domain]] of $f$. >$\huge >\int_{-a}^{a} f(x)\d x = >2\int_{0}^{a} f(x)\d x >$ ### Proof $\huge \begin{align} \let f &: X \to X \\ \let a &\in X \\ f(x) &= f(-x) \\ \int_{-a}^{a}f(x)\d x &= \int_{-a}^{0}f(x)\d x + \int_{0}^{a}f(x)\d x \\ &= \underbrace{\int_{-a}^{0}f(x)\d x}_{ u =-x, \d u = -\d{x} } + \int_{0}^{a}f(x)\d x \\ &= \int_{(-(-a)}^{-0}-f(u)\d u + \int_{0}^{a}f(x)\d x \\ &= -\int_{a}^{0}f(u)\d u + \int_{0}^{a}f(x)\d x \\ &= \int_{0}^{a}f(u)\d u + \int_{0}^{a}f(x)\d x \\ \let x &= u \\ \let du &= \d x \\ &=\int_{0}^{a}f(x)\d x + \int_{0}^{a}f(x)\d x \\ &= \boxed{2\int_{0}^{a}f(x)\d x} \\ \end{align}$ ## Odd >[!abstract] Thereom >A function $f(x)$ is [[Odd]] if $f(-x) = -f(x)$ for all $x$ in the [[Domain]] of $f$. >$\huge >\int_{-a}^{a} f(x)\d x = 0 >$