deletions | additions       diff --git a/untitled.tex b/untitled.tex  index b726072..96df64a 100644  --- a/untitled.tex  +++ b/untitled.tex 
...
If f is continuous on $R_2$ and $\int_{R}f dA = 0$ for all smoothly bounded sets $R$, then $$ is identically zero.  (a)If (a) If  $f(a,b)=p>0$ then there is a disc $D$ of radius $r>0$ centered at $(a,b) $in which $f(x,y)> \frac {1}{2}p$ (b) If f is continuous and f(x,y) $f(x,y)  ≥ p1 p_1  > 0 0$  on a disk R $R$  then $\int_{R}f dA \geq p1(Area(R))$. p_1(Area(R))$.  $\int_{R}f dA = 0$for all smoothly bounded regions $R$, then $f$ cannot be positive at any point.  (d) $f$ is not negative at any point either.   (e) $f = 0$ at all points. 
...
(a) Because f is continuous at $(a, b)$, by the definition of continuity, there is $r > 0$ such that for all $(x, y)$ such that$ ||(x, y) − (a, b)|| < r$, we have $|f(x, y) − f(a, b)| < p/2$.Then we assume p > 0, so $p − p/2 < f(x,y) < p + p/2$ In particular, $f (x, y) > p/2$  (b) As $R$ is bounded, the closure of $R$ is closed and bounded. So we can apply the extreme value theorem which means $f$ is bounded on the closure of $R$. In particular, $f$ is bounded on $R$. $f$ is also integrable on $R$; in fact $\int_{R}f dA = 0$.  Apply the lower bound property, $\int_{R}f dA \geq p1(Area(R))$ p_1(Area(R))$  holds. (c) Suppose $f$ is positive at $(a, b)$.