# If a continuous function has integral zero then what can we say about the function

Let $f$ be a continuous function on $[0,1]$. Suppose $$\int_0 ^x f(t)=0$$ for all x in [0,1] Then $f(t)=0$ for all $t\in[0,1]$. Is this statement true? If false can u give me a counter example.

• It is as false as you can expect. Just observe that $$\int_0^1\cos\pi x\,dx=0$$ For the claim to be true there is lacking one very important condition on $\;f\;$ ... – DonAntonio May 12 '18 at 18:23
• No, a function symmetric around the point $(x/2,0)$ would also do the job. And there are more counterexamples – Peter May 12 '18 at 18:24
• Is $x$ fixed, or is the integral 0 for all $x\in [0,1]$? – Alex R. May 12 '18 at 18:24
• If you mean this to hold for all $x$ in the interval, then it is true. Sketch: Suppose $f(x_0)>0$. By continuity suppose we have $\epsilon>0$ with $f(x)>0$ for $x\in [x_0-\epsilon, x_0]$. Then $\int_0^{x_0-\epsilon}f(t)\,dt$ and $\int_0^{x_0}f(t)\,dt$ can't both be $0$. But, really, your question is not clear. – lulu May 12 '18 at 18:29
• I apologize ...yes this holds for all x – Normal May 12 '18 at 18:56

Suppose $F(x)=\int_0^x f(t)\,dt= 0$ for all $x\in [0,1].$ Then obviously $F'(x)=0$ for all $x.$ But since $f$ is continuous, the FTC gives $F'(x) = f(x)$ for all $x$. Thus $0=F'(x)=f(x)$ for all $x.$
If you meant that $\forall x\in[0,1] \int\limits_{0}^x f(t)dt$, then yes, it follows that $f(x)=0\ \forall x\in[0,1]$.
Let's assume that $f(y)>0$ for some $y\in[0,1]$. Then $f$ is strictly positive in some neighbourhood of $y$. Then there is $\delta$ such that $\int\limits_{y-\delta}^{y+\delta} f(t)dt>0$, so $\int\limits_{0}^{y+\delta} f(t)dt=\int\limits_{y-\delta}^{y+\delta} f(t)dt+\int\limits_{0}^{y-\delta} f(t)dt>0$, which leads to a contradiction. Analogously for $f(y)<0$.