If the integral of a function is positive then the function is positive?

Question

Let $f:[a,b] \longrightarrow \mathbb{R}$ be a integrable function. I know if $ f> 0 $ then $$\int_{a}^{b} f(x)\; dx >0.$$ The converse is true? That is, if $$\int_{a}^{b} f(x)\; dx >0$$ then $f>0$? I couldn't think of an example that makes it false the converse.

Answer 1

What about $\displaystyle\int_{-1}^3x\,\mathrm dx$, which happens to be equal to $4$?

Answer 2

Take any function that is negative. Say $-1$ from $0$ to $1$. Compensate by a larger positive area, say $+1$ from $1$ to $2$.

Answer 3

If you do not restrict $f$ to be non negative,take $f=2\chi_{(1,2]} - \chi_{[0,1)}$ on $[0,2]$.Otherwise,even if you restrict $f$ to be non negative,this is not always the case.For example,$f=\chi_{(0,1]}$ on $[0,1]$.If you want a general answer in terms of the Lebesgue integral,it's true that on a measure space $X$, if the integral of a non negative measurable function $f$ is zero,then $f=0$ almost everywhere.You can prove this quite easily with Markov's inequality.

Answer 4

What does $\int_{a}^{b} f(x)\; dx >0$ show?....It says that the area covered by above x-axis is more that area covered by the part below x-axis.

For eg. a function which is always positive will always lie above the x-axis, so the area above x-axis will be always more that below (which will be zero here).

eg. $f(x) = (x-3)^2 +2 $

But the fact that the integral is positive does not mean that the function is positive, as it could also mean that the area above the x-axis is more. So portion of the function could exist below graph in such a condition, provided that its area under is less that the part above.

eg. $f(x) = x(x-3)(x-5)$ from $x=0.5$ to $x=3.5$

Here the portion (1) which is above the x-axis is definitely larger than the portion (2) below x-axis. So the definite integral will be positive here, but notice how the function is also negative .