By fundamental theorem of calculus, show that $f$ is odd if $\int_{-s}^sf(x)dx=0$, $\forall s\in \mathbb{R}$, $f$ being continuous function.

Question

By fundamental theorem of calculus, show that $f$ is odd if $\int_{-s}^sf(x)dx=0$, $\forall s\in \mathbb{R}$, $f$ being continuous function.

Fundamental theorem of calculus:

1st FTC: If $f$ is continuous on $[a,b]$ then the function $F$ defined by

$F(x)= \int_{a}^{x} f(t)dt$ for $a \le x \le b$ is continuos on $[a,b]$ and differentiable on $(a,b)$ , and $F'(x)=f(x)$.

2nd FTC: If $f$ is continuous on $[a,b]$, then $\int_{a}^{b}f(x)dx=F(b)-F(a)$, where $F'=f$.

How to prove my problem using FTC, please help.

Answer 1

$F(s)-F(-s)=0$, differentiating both sides, $f(s)+f(-s)=0$.

\begin{align*} F(s)&=\int_{a}^{s}f(t)dt,\\ F(-s)&=\int_{a}^{-s}f(t)dt=-\int_{-s}^{a}f(t)dt\\ F(s)-F(-s)&=\int_{a}^{s}f(t)dt+\int_{-s}^{a}f(t)dt=\int_{-s}^{s}f(t)dt. \end{align*}

Answer 2

$$ \int_{-s}^{s} f(x)dx = 0 \implies \int_{-s}^{0} f(x)dx + \int_{0}^{s} f(x) dx= 0 $$ You may use FTC here : Let $$ F(s) = \int_{0}^{s} f(x) dx $$, so we have $$\int_{-s}^{0} f(x)dx + \int_{0}^{s} f(x)dx = 0 \implies -F(-s)+F(s)=0 $$ $$ F(s)=F(-s) $$ You may continue from here..

Answer 3

Correct me if wrong.

Let $f$ be continous on $\mathbb{R}:$

$F(s):=\displaystyle \int_{a}^{s} f(x)dx$, $F$ differentiable.

$\displaystyle \int_{-s}^{s} f(x)dx = 0$ implies

$F(s)=F(-s)$, $s \in \mathbb{R}.$

Since $F(s)$ is even, its derivative is odd:

$F'(-s)=- F'(s).$

Now by FTC:

$F'(s)=f(s)$, and

$F'(-s)= f(-s)$;

Hence: $f(-s)= -f(s)$, odd.