# If f is continuous on [a,b], f(x)>0, and f(x0)>0 for some x0 in [a,b]

1. Aug 22, 2008

### Nomialists

First of all, hello everyone, this is my first post so I am not sure if this the right place to post this question.

I am wondering if anyone can help me understand this question better.

The question goes as: if f is continuous on [a,b], f(x)>0, and f(x0)>0 for some x0 in [a,b], prove that $$\int^{b}_{a}f(x)dx>0$$.
(Hint: By continuity of f, f(x)>1/2f(x0)>0 for all x in some subinterval [c,d]. Use a) and b) steps )

a) 1) Assume f is integrable on [a,b]. Prove:
If f(x)>=0 on [a,b] then $$\int^{b}_{a}f(x)dx>0$$.

Proof:
Since every approximating sum $$\sum^{n}_{k=1}f(x)\Delta x>0$$
then $$\int^{b}_{a}f(x)dx>0$$

2) If m<=f(x)<=M for all x in [a,b] then $$m(b-a)<=\int^{b}_{a}f(x)dx<= M(b-a)$$

Proof:
$$\int^{b}_{a} m dx <= \int^{b}_{a} f(x) dx<= \int^{b}_{a}Mdx$$
$$\int^{b}_{a} m dx = m \int^{b}_{a} 1dx = m(b-a)$$ and
$$\int^{b}_{a} M dx = M \int^{b}_{a} 1dx = M(b-a)$$ then
$$m(b-a)<=\int^{b}_{a}f(x)dx<= M(b-a)$$

b) If a<c<b then f(x) is integrable on [a,b] iff it is integrable on [a,c] and [c,b]. Moreover if f is integrable on [a,b].

$$\int^{b}_{a}f(x)dx = \int^{c}_{a}f(x)dx + \int^{b}_{c}f(x)dx$$

I don't understand why I even need to use b) or even the second part of a). Since f is continuous then it's integrable [a,b] so I can simply replicate the proof of a) 1 to solve this one. I got this question out of Schaum's Outline of Calculus. I just don't know why the book even mention f(x)>1/2f(x0). a) and b) are previous questions to this problem.

Any help to understanding this will be much appreciated.

Thank You

2. Aug 22, 2008

Surely the second condition should be $f(x)\geq 0$, and not $f(x)>0$? Otherwise, you can simply observe that $f(x)>c>0$, and so $\int_a^b f(x) dx > \int_a^b c dx = c(b-a)>0$.

3. Aug 23, 2008

### d_leet

That doesn't follow from f(x)>0 consider the function f(x)=1/x2 then f(x)>0 everywhere but there is no c such that f(x)>c everywhere.

And the theorem is false if we allow $f(x)\geq 0$ since f(x)=0 satisfies that condition, but does not have a nonzero integral.

4. Aug 23, 2008

### mrandersdk

f(x)=0 doesn't satisfy f(x_0) > 0 for some x_0 in [a,b]. And f(x)=1/x^2 have such a c, fx.

1/(a^2+epsilon)

5. Aug 23, 2008

On a finite interval $[a,b]$ there is such a $c$. It is $c = b^{-2}$.

No, the theorem is still true, because of the other conditions ($f(x_0)>0$ for some $x_0 \in [a,b]$ and $f(x)$ continuous). The idea is to show that there is a measurable difference between two continuous functions, if you know they differ at a single point (which itself has measure zero).

It seems to me that what they want you to do in this proof is show that there's always some component of the integral that is positive. I think the same $>$ typo has propagated to a) 1); they should all be $\geq$ as well. Then in 2) you show that if $f(x)>0$, $\int_a^b f(x)dx > 0$ as well. Finally, part 3 reminds you that you can partition $[a,b]$ and then add the resulting integrals. So, partition $[a,b]$ according to whether $f(x)=0$. Then show that the $f(x)>0$ partition (i.e., the one containing $x_0$) must have a nonzero integral.

6. Aug 23, 2008

### mrandersdk

exactly what I tried to say that, but put a instead of b, which is ofcause not true, yours is right.