Hello, Runty!
I have an intuitive (graphic) proof . . .
Prove that, if $\displaystyle f(x)$ is continuous on $\displaystyle [a,b]$, and: .$\displaystyle \int_{a}^{b} f(x) dx \:=\: 0$
. . then $\displaystyle f(x)\,=\,0$ for all $\displaystyle x \in [a,b].$
A hint provided was to prove this via contradiction. Suppose $\displaystyle f(x)$ is not equal to 0 for all $\displaystyle x \in [a,b].$
. . Then $\displaystyle f(x)$ is not the $\displaystyle x$axis.
There are three cases:
. . (1) $\displaystyle f(x)$ is above the xaxis on the interval $\displaystyle [a,b]$
. . (2) $\displaystyle f(x)$ is below the xaxis on the inverval $\displaystyle [a,b].$
. . (3) $\displaystyle f(x)$ crosses the xaxis on the interval $\displaystyle [a,b].$
For case (1), we have this graph: Code:

 *
 *:
 *:::
 *:::::
 *:::::::
 :::::::
    +    +  
 a b
$\displaystyle \int^b_af(x)\,dx$ represents the area under the curve,
. . . which is evidently not zero.
For case (2), we have this graph: Code:

 a b
  +  +    +  
 :::::::
 *:::::::
 *:::::
 *:::
 *:
 *
$\displaystyle \int^b_af(x)\,dx$ represents the area above the curve,
. . . which also is not zero.
For case (3), we have this graph for $\displaystyle f(x)$: Code:

 *
 *
 *
 *
  +  +    *   +   
 a * b
 *
 *
The the graph of $\displaystyle f(x)$ looks like this: Code:

 *
 :* *
 :::* *:
 :::::* *:::
    +    *    +   
 a b

$\displaystyle \int^b_af(x)\,dx$ represents the area under the curve,
. . . which again is not zero.
[This may not be considered a a satisfactory proof.]