1
$\begingroup$

For $a, suppose that $f$ is continuous on $[a,b]$ and $\int_{a}^{b}f(x)dx=0$. Prove that there is at least one number in $(a,b)$ such that $f(c)=0$.

I attempted this

Let $c$ be any number in $(a,b)$

$\int_{a}^{b}f(x)dx=0$ implies $\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx=0$ implies $\int_{a}^{c}f(x)dx=-\int_{c}^{b}f(x)dx$

Since $a, the equation tells us that the interval before $c $ is of opposite polarity and equal to the interval after $c$. That means the function goes from (1). either negative to positive or (2).positive to negative or the (3).function is zero.

(3) If the function is zero, we are done as f(c)=0.

(1),(2) That is, either $f(a)>0 $ and $f(b)<0$ or $f(a)<0 $ and $f(b)>0$, and since $f(x)$ is continuous, by the Intermediate Value Theorem, there exists a $c$ such that

$f(c)=0$

I am very unsure if I would get my marks for showing this. Is it a correct proof? I am also looking for more elegant proofs. Thanks in advance!

  • 0
    wow! I guess that counterexample killed it! damnnn...2012-10-29

3 Answers 3

5

Put

$ F(x) = \int_{a}^{x}f(x)dx $

and use Rolle's theorem.

1

Use the mean value theorem on the function $g(x) = \int_a^{x}f(t)dt$. This problem is a special case of the mean value theorem for integrals.

1

If $f$ is never $0$ in the $[a,b]$ interval, then, by continuity, either $f>0$ or $f<0$. Wlog suppose the former. Now, by $[a,b]$ being compact, $f$ has a minimum $m$ in $[a,b]$, which is also $>0$, hence $\int_a^b f \ge m(b-a) >0$.