Integral being less than or equal to zero.

Question

I'm having a bit of trouble understanding this question:

1. If $f(x) \leq 0$ for all $x \in [a, b]$ then $\int_a^b f(x) dx \leq 0$.
2. If $\int_a^b f(x) dx \leq 0$ then $f(x) \leq 0$ for all $x \in [a, b]$.

Are the above statements True of False? Justify your response.

I believe it is asking if the area under the graph between $a$ and $b$ is negative then $f(x)$ will also be below zero on the graph and vice versa. However I'm still not sure how to justify my response.

Answer 1

The first is true cause all the upper sums are negative.

the second is false. take the example:

$$f:[0,1]\to\mathbb R $$ $x \mapsto -7$ if $x\neq 0$ and $f (0)=+1$. then $$\int_0^1 f=-7$$ but $f (0)>0$. Or

$$\int_{\frac {\pi}{2}}^\pi \sin (x)dx=-1$$

but $\sin (\frac {\pi}{2})=1>0$.

Answer 2

To justify 1.: If you have $g, g': [a,b] \longrightarrow \mathbb{R}$ such that $g\leq g'$ (pointwise), then how do the integrals relate (you should have had a theorem like this, it's a basic property of integrals)?

For 2., a counterexample would provide a proof.
A nonzero function $f: [a,b] \longrightarrow \mathbb{R}$, such that its integral splits up into two integrals of inverse orientation of same absolute value (i.e., same area between the negative part of the graph and the $x$-axis as between the nonnegative part and the $x$-axis) will do.
There are very simple ones (note that for a counterexample, you are free to choose $a$ and $b$).

Salahamam_'s counterexample will do, but it doesn't have much educational value...