1. Does this necessarily follow?

Assume that f and g are continuous on [a,b] and

$\displaystyle \int_a^b f(x) dx > \int_a^b g(x) dx$

Does it necessarily follow that $\displaystyle f(x)>g(x)$ for all $\displaystyle x\in[a,b]$?

2. No
Take for instance f(x) = 1/2 and g(x) = x over [-1,1]

3. No, put $\displaystyle f(x)=\frac1{1+x^2}$ and $\displaystyle g(x)=x$ for $\displaystyle 0\le x\le1.$

But $\displaystyle \frac1{1+x^2}\not>x.$ (At least not always.)

(Anyway, the converse is true.)

4. $\displaystyle If\ f(x)>g(x)\ for\ all\ x\ over\ an\ interval,$

$\displaystyle then\ \int{f(x)dx}>\int{g(x)dx}\ certainly,\ over\ the\ interval.$

The converse is not true however.
Similar to the graph given by running-gag,

$\displaystyle f(x)=4x\ from\ x=0\ to\ x=1$

when integrated, gives 2.

$\displaystyle g(x)=x+1$

when integrated over the same interval gives $\displaystyle \frac{3}{2}$

$\displaystyle f(x)<g(x)\ from\ x=0\ to\ x=\frac{1}{3}$

but

$\displaystyle f(x)>g(x)\ from\ x=\frac{1}{3}\ to\ x=1.$