Prove that there exists a point $x_0∈ [a,b]$, where $f(x)$ is continuous and $f(x_0)>0$, then $∫^a_bf(x)dx>0$

Question

I have that the function $f(x)$ is integrable and $f(x)>0$ in the interval $[a,b]$ I need to prove that there exists a point $x_0 ∈[a,b]$, where $f(x)$ is continuous and $f(x_0)>0$, then $∫^a_bf(x)dx>0$

Any ideas?

Answer 1

Your question is a little messy in terms of its statements and phrasing, so I'll demonstrate two separate and possible cases :

First case :

If you have the facts : $$f(x) \space\text{continuous and integrable on} \space [a,b] \space \text{and} \space f(x_0) > 0 \space \text{for} \space x_0 \in [a,b]$$

then you're missing out on one fact, that $b-a >0$.

To see that, take the function :

$$g(t) = \int_a^t f(x)dx$$

The function $f(x)$ is continuous which means that $g(x)$ will be continuous in such an interval and of course differentiable too. Taking the interval $[a,b]$ which is the one that is nicely set and defined for $f(x)$ via the exercise, the Mean Value Theorem can be applied for $x_0 \in \mathbb (a,b)$ :

$$g'(x_0)=\frac{g(b)-g(a)}{b-a} = \frac{\int_a^bf(x)dx-\int_a^af(x)dx}{b-a} = \frac{\int_a^bf(x)dx}{b-a}$$

$$ \Rightarrow $$

$$\bigg[\int_a^tf(x)dx\bigg]'_{t=x_0} = \frac{\int_a^bf(x)dx}{b-a} \Rightarrow f(x_0) = \frac{\int_a^bf(x)dx}{b-a} \Leftrightarrow \int_a^bf(x)dx=f(x_0)(b-a)$$

Now, if $f(x_0)>0$ from the facts given. If $b-a>0$, then we have proved that :

$$\int_a^b f(x)dx>0$$

Second case :

If you have the facts : $$f(x) \space\text{continuous and integrable on} \space [a,b] \space \text{and} \space \int_a^b f(x) > 0$$

then one can prove by taking the same function as above :

$$g(t) = \int_a^t f(x)dx$$

and following a similar path by applying the Mean Value Theorem, that there exists $x_0 \in [a,b] : f(x_0) > 0$.

Note : ("Shout out" to @DougM as he also mentioned the Mean Value Theorem approach in the comments as I was writing the answer)