To find  \(\int_{a}^{b}\left(f(x)-g(x)\right)dx\)  requires careful observation of the two shaded regions shown.

The green region shows  \(f(x)\ge g(x)\), while the blue region shows  \(g(x)\ge f(x)\).   The changing point is at  \(x=c\).   Therefore, two different calculations need to be made: $$\color{green}{\int_{a}^{c}\left(f(x)-g(x)\right)dx} + \color{blue}{\int_{c}^{b}\left(g(x)-f(x)\right)dx}$$ Finding the intersection points, aka  \(c\), is sometimes a challenge!   Also, maybe the values of  \(a\)  and  \(b\)  may not be given either.

Attempting to find the area bounded by the functions yields:  \(\int_{0}^{9}\left(f(x)-g(x)\right)dx\).   However, notice that from  \(x=0\)  to  \(x=4\)  the red graph is the top boundary and also the bottom boundary, or  \(f(x)=g(x)\).   This would produce  \(\int_0^4\left(f(x)-f(x)\right) dx=0\)  in the area formula, causing a mistake.

Instead, looking horizontally, the region is bounded "above" by the green line and "below" by the red function.   By rewriting the functions to use the  \(y\)-variable, the area formula changes to:  \(\int_{c}^{d}(f(y)-g(y))dy\)  where  \(c\)  and  \(d\)  are the  \(y\)-coordinates.   The area wold then be found by: $$\int_{1}^{6}\left(\color{green}{g(y)}-\color{red}{f(y)}\right)dy$$ The key piece here is to write each function in terms of  \(y\).