Integral equal zero implies that f(x) has a root

Question

I have $\int_{-1}^1 f(x) dx=0$ and need to show that $f(x)$ has a root in $[-1,1]$. $f(x)$ is continuous.

Answer 1

Since your function is continuous you may use the mean Value theorem for integrals to get $$\int_{-1}^1 f(x) dx= 2f(c)$$ for some $c\in [-1,1]$

That implies$$ f(c)=0 $$ for some $$c\in [-1,1].$$

Answer 2

This can be proven using the Mean Value Theorem, and I will give an intuition of it here.

The case is trivial for the zero function, so consider when $f$ is not.

Then in $[-1,1]$, for the condition to hold, we must have at least one part of $f$ above the $x$-axis and at least one below since $f$ is continuous.

For example, we can write $$\int_{-1}^1f(x)\,dx=\overbrace{\int_{-1}^{x_1}}^{\text{above}}+\overbrace{\int_{x_1}^{x_2}}^{\text{below}}+\cdots+\overbrace{\int_{x_{k-1}}^{x_k}}^{\text{above}}+\overbrace{\int_{x_k}^1}^{\text{below}}$$ and it becomes clear that $x_1,\cdots,x_k$ are the roots.

Answer 3

If $f$ had no root the, since it is continuous, you would have $(\forall x\in[-1,1]):f(x)>0$ or $(\forall x\in[-1,1]):f(x)<0$. Therefore, again because $f$ is continuous,$$\int_{-1}^1f(x)\,\mathrm dx>0\text{ or }\int_{-1}^1f(x)\,\mathrm dx<0.$$

Answer 4

You can use the continuity of f(x) on [-1, 1] to prove

$$f(c)=0$$ for some $$c\in [-1,1]$$

If $$f(x)\ne 0 $$ for all $x\in [0,1]$, then by continuity of f(x) we conclude that either $f(x)>0$ for all$ x\in [-1,1]$ or $ f(x)>0$ for all $x\in [-1,1]$

In any of these cases we will get

$$\int_{-1}^1 f(x) dx \ne 0$$

Thus $f(c)=0$ for some $c\in [-1,1]$