Show that the $f(x)$ is an odd function

Question

A function $f$ is continuous on $\mathbb{R}$ and $\displaystyle\int_{-x}^xf(t)\,\mathrm{d}t = 0$ for all $x\in \mathbb{R}$. Show that $f$ is an odd function on $\mathbb{R}$. How to show this by using Riemann Integral concepts.

Answer 1

HINT: $F(x)$ is even (why?). Prove that the derivative of even function is odd.

Answer 2

Just use the fundemental theorem of calculus.

Let $F(x) = \int_{0}^{x} f(t) dt$

Then we have that $F(x) - F(-x) = 0$ Taking derivatives of both sides gives us that $F'(x) + F'(-x) = 0$

Answer 3

Let $x \in \mathbb{R}$. \begin{align}\newcommand{\dx}[1]{\,\mathrm{d}{#1}} 0 &= \int_{-x}^x f(t)\dx{t} &&\\ &= \int_{0}^x f(t)\dx{t} + \int_{-x}^0 f(t)\dx{t}&&\text{(additivity with respect to interval of integration)}\\ &= \int_{0}^x f(t)\dx{t} + \int_{0}^x f(-t)\dx{t}&&\text{(substitution $t \mapsto -t$)}\\ &= \int_{0}^x \big(f(t)+f(-t)\big)\dx{t}&&\text{(linearity of the integral)}\\ \end{align}

Taking the derivative $\frac{d}{dx} $of both sides yields $f(x) = -f(-x)$.