Fundamental Theorem of Calculus application for $f(x)\geq 0$ Can anybody help me with how to solve the following question using the fundamental theorem of calculus? I'm a bit confused...
If $f$ is a continuous function on $[a, b]$ and  $f(x)\geq 0$ for all $x\in [a, b]$. Show that if $\int\limits_a^b f(x)dx = 0$ then $f(x) = 0$ for all $x\in [a, b]$.
 A: we will prove by contradiction.
suppose $f$ is not identically zero in $[a, b].$  then there is a point $c \in [a, b]$ such that $f(c) > 0.$  by continuity, there is $\delta > 0,$ such that $f(x) \ge \frac12 f(c) , \text{ for all } x \in [c-\delta, c+ \delta] \cap [a, b].$  this means $$\int_a^b f(x) \, dx \ge \frac12 \delta f(c) > 0$$ contradicting the hypothesis $\int_a^b f(x) \, dx = 0.$
A: For fun, we will use the Fundamental Theorem of Calculus, even though using it gives  a proof that is more complicated, and less intuitive geometrically, than the argument used by abel.
Suppose to the contrary that $f(c)\gt  0$ at some point $c$ in our interval. Let $F$ be an antiderivative of $f$. Then $F'(c)\gt 0$. By continuity, there is an interval in which $F'\gt 0$. Then $F$ is strictly increasing in that interval, contradicting the fact that $F(b)-F(a)=0$.
A: Consider the  function $F(x)={\int}_a^xf(t)dt$. Then, by the fundamental theorem, $F'(x)=\frac{d}{dx}(0)=0=f(x)$.
EDIT: To show that ${\forall}x{\in}[a,b]\,F(x)=0$, use the fact that $f(t)\geq0$, so ${\forall}x\in[a,b](0{\leq}F(x){\leq}F(b)=0)$, so $F(x)=0$.
A: Assume there is a point $x_0\in [a,b]$ such that $f(x)\gt 0$. Thanks to the continuity of $f$ there is an interval $[c,d], c\neq d$ containing $x_0$ such that $f(x_0)\gt \frac{f(x_0)}{2}$.
Now decompose the integral the integral over $\int_c^d f(x)dx\gt \frac{f(x_0)}{2}(d-c)\gt 0$ while the integrals over $[a,c]$ and $[d,b]$ are non negative and therefore $\int_a^b f(x)dx\gt \frac{f(x_0)}{2}(d-c)\gt 0$ A contradiction
A: The fundamental theorem of calculus tells us $$\int_a^b f(x)\,dx = F(b) - F(a), where\,  F'(x) = f(x)$$
so therefore $$F(b) - F(a) = 0$$
This gives us that $$F(b) = F(a)$$
If F(x) is constant then F'(x), or f(x) must be zero.
EDIT: 
A more mathematical way to make that last statement
$$f(x) = F'(x) = \frac{F(b) - F(a)}{b-a} = 0/(b-a) = 0$$
A: Hint: You can prove this by contraposition using the definition of lower Riemann sums. Assume that $f(x)$ is positive somewhere, use the definition of continuity and then find a partition that can prove the contrapositive.
