There is a continuous function not identically zero such that There is a continuous function  $f:[a,b] \longrightarrow \mathbb{R}$ not identically zero in $  I=[a,b]$  such that $$\int_a^b|f(x)| \, dx=0 ~~~~?$$
I have problems
 A: This is false. If $f$ is not identically zero on the interval, then there is some point $x_0$, $m>0$ and a neighborhood $(x_0-\frac{1}{2}\delta,x+\frac{1}{2}\delta)$ for which $|f(x_0)| \ge m > 0$. You can prove this with a simple $\varepsilon$-$\delta$ argument. Thus the area under $|f(x)|$ will be greater than or equal to $m\delta$.
If $f$ is relaxed to be almost everywhere continuous, then it's true that you could have a function that is not identically zero but the integral of $|f|$ is zero.
A: Define
$$
F:[a,b]\to\mathbb R,\ x\mapsto\int_a^x|f(t)|dt.
$$
Since $|f|$ is continuous, $F$ is a (differentiable) primitive of $|f|$ by the fundamental theorem of calculus. By assumption, $F=0$ in $[a,b]$. Thus, $0=F'=|f|$ in $[a,b]$ and hence $f=0$ in $[a,b]$.
A: No, the integral of a positive valued continuous function is 0 if and only if it's identically zero.
A: Let $\phi(x) = |f(x)|$. If $\int \phi = 0$, then the lower Riemann sum satisfies $L(\phi, \Pi) = 0$ for any partition $\Pi$. In particular, for each interval $I$ in the partition, we must have some $x \in I$ such that $\phi(x) = 0$.
Since $[a,b]$ is compact, $\phi$ is uniformly continuous, hence for any $\epsilon>0$, there is some $\delta>0$ such that $|x-y| < \delta$ implies $|\phi(x)-\phi(y)| < \epsilon$.
Choose $\epsilon>0$, and choose any partition $\Pi$ with mesh size less than $\delta$. Each interval $I$ in the partition contains a point $x$ such that $\phi(x) = 0$, hence uniform continuity shows that $\sup \phi(I) < \epsilon$. Hence $\sup \phi([a,b]) < \epsilon$. Since $\epsilon>0$ was arbitrary, we have $\sup \phi([a,b]) = 0$. Hence $f(x) = 0$ for all $x \in [a,b]$.
