Let $\varphi\colon[-1,1]\to \mathbb R$ be an odd step function.Prove that: $\int_{-1}^1\! \varphi(t)\, dt = 0$
Thanks!
Let $\varphi\colon[-1,1]\to \mathbb R$ be an odd step function.Prove that: $\int_{-1}^1\! \varphi(t)\, dt = 0$
Thanks!
Let $f$ be an odd integrable function, that is, $f(x)=-f(-x)$ for all $x\in\mathbb{R}$. Then:
$\int_{-R}^R f(x) dx=\int_{-R}^0f(x)dx+\int_0^Rf(x)dx=-\int_0^Rf(x)dx+\int_0^Rf(x)dx=0$