When $f(0) = 0,$ is it always true that $G(0) = 0,$ where $G$ is the antiderivative of $f$?

Question

I have a hunch that it is, but it would be nice if somebody could confirm / disprove it for me.

Thank you.

Edit Is it when the constant of integration is equal to zero?

Answer 1

No.

Take $f(x) = xe^{-x^2}$. Clearly, $f(0) = 0$.

The integral function of $f(x)$ is:

$$G(x) = \int_{-\infty}^{x} te^{-t^2} dt = \left.-\frac{1}{2}e^{-t^2}\right|_{t=-\infty}^{t=x} = -\frac{1}{2}e^{-x^2}$$

In this case, $G(0) = -\frac{1}{2} \neq 0.$

Answer 2

No, this is not true in general. Let $F(x)$ be a anti-derivative of $f(x)$ then $$\int f(x) \, \mathrm{d}x = F(x) + k$$ now, even if $F(0) = 0$ there's still the $+ k$ constant to contend with.

An example would be $f(x) = x\exp x$ then we have $$F(x) = \int x\exp x \, \mathrm{d}x = e^x (x-1) + k$$ Now, even if $k = 0$, then $F(0) = -1$ whilst $f(0) = 0$.

Answer 3

As I commented, another counterexample is given by $f(x)=e^x-e^{-x}$, whose antiderivative is $F(x)=e^x+e^{-x}+C$.

Note that $f(0)=e^0-e^0=0$, and $F(0)=e^0+e^0+C$, so $F(0)=0\iff C=-2$.

Variations of the coefficients of the exponentials or the exponents would force you to choose a different value for $C$:

• $f(x)=2e^x-e^{-x}$, with antiderivative $F(x)=2e^x+e^{-x}+C$. Then $F(0)=0 \iff C=-3$
• $f(x)=e^{2x}-e^{-x}$, with antiderivative $F(x)=\frac{1}{2}e^{2x}+e^{-x}+C$. Then $F(0)=0 \iff C=-\frac 32$.

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However, your statement holds for polynomial functions:

Let $f(x)=a_0+a_1x+\cdots a_nx^n$.

Since we want $f(0)=0$, the constant term $a_0=0$, so $f(x)=a_1x+\cdots +a_nx^n$.

The antiderivative of $f(x)$ is $F(x)=\frac{a_1}{2}x^2+\frac{a_2}{3}x^3+\cdots+ \frac{a_n}{n+1}x^n+C$, so requiring that $C=0$ does ensure that $F(0)=0$.

Answer 4

No. The simplest example I can think of? $$f(x) = x$$ Then $G(x) = \frac{x^2}{2} + C$, which is not guaranteed to be always $0$.