If $g(x)$ is an even function with $f(0)=0$, then how can we deduce that $f(x)$ is an odd function?

Question

Let $f(x)=\int_0^x g(t)\,\mathrm dt$, where g is a non-zero even function. If $f(x + 5) = g(x)$, then $\int_0^xf(t)\,\mathrm dt$, equals:$$\int_{x+5}^5g(t)\,\mathrm dt,\quad2\int_5^{x+5}g(t)\,\mathrm dt,\quad\int_5^{x+5}g(t)\,\mathrm dt,\quad5\int_{x+5}^5g(t)\,\mathrm dt?$$(One out of these four is correct)

The book gives the hint that $f(0)=0$, so, $f(x)$ is an odd function. After this information, I am able to solve the question. But I wonder how could $f(0)=0$ lead to $f(x)$ being an odd function. Seems like it has something to do with $g(x)$ being an even function, but not able to figure it out. (Though if I replace $x$ by $-x$ in the given exxpression, I am able to see that $f(x)$ is indeed odd.)

Answer 1

The derivative of an odd function is even. Always. This can be shown by the chain rule.

But an antiderivative of an even function may fail to be odd because of the integration constant.

E.g. $x^3$ odd implies $3x^2$ even. But $3x^2$ even does not imply $x^3+5$ odd. You need an extra condition, namely $f(0)=0$.

Answer 2

Replacing $x$ by $-x$, we get, $$f(-x) = \int_0^{-x}g(t)dt.$$ With the substitution $t = -u$, we get $$f(-x) = -\int_0^{x}g(-u)du.$$ As $g$ is even, we have that $g(-u) = g(u)$, this gives us $$f(-x) = -\int_0^{x}g(u)du = -f(x).$$

Hence, $f$ is odd.

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Without $g$ being even, it is not necessary that $f$ is odd, btw. Take $g(x) = x$ as a counterexample. However, you still have $f(0) = 0$ in this case. Thus, you are correct in the fact that $f(0) = 0$ does not let you conclude that $f$ is odd.