# Verification: If $\int_a^bf(x)dx=0$ then there is $x_0\in [a,b]$ for which $f(x_0)=0$

Old exam question: Let $$a,b\in\mathbb{R} : a and let $$f$$ be a continuous function, $$[a,b]\rightarrow\mathbb{R}$$.

Show that if $$\int_a^bf(x)dx=0$$ then there is $$x_0\in [a,b]$$ such that $$f(x_0)=0$$

Hint: $$f$$ can only have a sign, if $$f$$ has no roots.

My approach:$$f$$ is continuous so by the fundamental theorem there is $$F$$ such that $$\int_a^bf(x)dx = F(b)-F(a)=0$$ and $$F'=f$$

Then by the intermediate value theorem for derivatives we have $$0=\frac{F(b)-F(a)}{b-a}=F'(\zeta)=f(\zeta)$$ for some $$\zeta\in(a,b)$$

Now for my question: why the hint? I'm afraid I may have misunderstood the question as I don't even know how to use the hint.

• @MatthewDaly Thanks, edited – Ruben Kruepper Aug 14 at 10:26
• By the hint it suffices to consider $f>0$ or $f<0$ everywhere, which is a lot easier. – trisct Aug 14 at 10:36
• Your solution is OK. – Dr Zafar Ahmed DSc Aug 14 at 13:21

## 1 Answer

Assume $$a. Proof suggested by the hint: If $$f$$ has no zeros then it has the same sign throughout the interval. Let us say $$f(x) >0$$ for all $$x$$. Since $$f$$ is continuous it attains its minimum value $$m$$ so $$m >0$$. This gives $$\int_a^{b}f(x)dx >m(b-a)$$ so the integral cannot be $$0$$.

• Advantage with this proof: no derivatives used. – Kavi Rama Murthy Aug 14 at 10:28
• Thank you very much, sir, very elegant! For peace of mind: was mine wrong or just lengthy? – Ruben Kruepper Aug 14 at 10:30
• Nothing wrong at all with your proof. – Kavi Rama Murthy Aug 14 at 10:31