So I probably didn't understand it at first, but from what I learned, integrating two functions that are both below the x-axis would mean that we would do $\int-f(x)-g(x)$ instead of just $\int f(x)-g(x)$. Not sure if thats the case because I just had something where both the functions were under the x-axis on the given interval, and we still had to go by the $\int f(x)-g(x)$ order to solve it correctly.
Would appreciate some clarification on this matter.
Thanks in advance.
On
If you want the area between the curves $y=f(x)$ and $y=g(x)$ on the interval $[a,b]$, and if $f(x) \ge g(x)$ on $[a,b]$, then the signs of $f,g$ don't matter. The vertical strips have heights $f(x)-g(x)$ (bigger $y$-value minus the smaller $y$-value), so the area is $$\int_a^b f(x)-g(x)\,dx$$ without any reference to where the curves are relative to the $x$-axis.
As another way to see it, we could shift both curves up by some positive constant $c$, big enough so that, on the interval $[a,b]$, the shifted curves are both above the $x$-axis. But the area of the shifted region is the same as that of the original region, hence the area, still assuming $f(x) \ge g(x)$ on $[a,b]$, is equal to $$\int_a^b \bigl(f(x)+c\bigr)-\bigl(g(x)+c\bigr)\,dx$$ which is algebraically the same as $$\int_a^b f(x)-g(x)\,dx$$ If the curves $y=f(x)$ and $y=g(x)$ cross each other on the interval $[a,b]$, you have to worry about the subintervals of $[a,b]$ where one function has $y$-values greater than the other. The area between the curves is then the sum of the appropriate integrals for each of the subintervals.
If you really want to have the size of the area between the two curves, the formula to go with is $$\int_a^b\left|f(x)-g(x)\right|dx$$ Where $f(x)\geq g(x)$, this translates to $$\int f(x)-g(x) dx$$ Where $f(x)\leq g(x)$, it translates to $$\int g(x)-f(x) dx$$ But it never will be $$\int -f(x)-g(x) dx$$ if you look for the area between the two curves.