Proof involving Riemann integration/fundamental theorem of calculus Suppose $f$ is a continuous function on $[x,y]$ and for all $x\le a \le b \le y$,  $\int^b_a f \,dx = 0$. Then why must $f(u)=0$ for all $u\in[a,b]$?
 A: Why you don't just say:
$\frac{d}{dz}\int^z_x f(t) dt =$$ f(z) = 0$, 
 for all $z \in [x,y]$?
A: As has been pointed out, since $f$ is continuous you can use the fundamental theorem.
Another way is this: Suppose $f(c)>0$.  Then $f>0$ everywhere in some interval containing $c$ (because of continuity), so the integral of $f$ over that integral is positive.  A similar thing can be done if $f(c)<0$.
This is also essentially equivalent to at least one other posted answer, but I prefer to say it more simply.
A: Suppose there exists $x_0\in (a,b)$ such that $f(x_0)>0.$ Since $f$ is continuous there exits $r>0$ such that $f(x)>0$ for all $x\in [x_0-r,x_0+r]$ and $[x_0-r,x_0+r]\subset [a,b].$ (Note that by Weierstrass theorem, there exists $k>0$ such that $f(x)\ge k,\forall x\in [x_0-r,x_0+r]$). Now, we have
$$0\underbrace{=}_{\mathrm{assumption}}\int_{x_0-r}^{x_0+r}f(x)\,dx \underbrace{>}_{f\ge k} 2rk>0.$$ This gives us a contradiction. 
Arguing in a similar way if you assume $f(x_0)<0$ we get also a contradiction.
So, $f\equiv 0$ on $(a,b).$ By continuity it is $f\equiv 0$ on $[a,b].$
A: If $f$ is not identically zero on $[a,b]$ one can choose a subinterval $[c,d]$ on which either $f \ge k_1 >0$ or else $f \le k_2 <0.$ Now integrate $f$ against the characteristic function of $[c,d]$ (or if you want to use continuous functions, adjust near the endpoints).
Support for the continuity implying "one can choose" etc.:
Suppose $f(x_0)>0$ and let $k=f(x_0)/2.$ Then by continuity of $f$ at $x_0$ choose $\delta$ for which $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<k.$ Then it will hold that for $x \in [x_0-\delta/2,x_0+\delta/2]$ we have $f(x)\ge k.$ Finally using this interval as the interval $[c,d]$ one would by the assumptions get the integral $0$ but since $f\ge k$ we in fact have the integral at least $k(b-a).$
