Answer
$\frac{4}{3}$
Work Step by Step
Setup the integration to find the area between the two functions.
over the interval [0,4], $f(x) \gt g(x)$
$\int_0^4 [ (\sqrt x +3)-(\frac{1}{2}x +3)]dx$
$\int_0^4 (\sqrt x +3 -\frac{1}{2}x -3)dx$
$\int_0^4 (\sqrt x -\frac{1}{2}x)dx$
$[ \frac{2}{3} x^{\frac{3}{2}} - \frac{1}{4}x^2]_0^4$
$ (\frac{16}{3} -4) -0$
$\frac{4}{3}$
