$(1$ point$)$ Consider the area between the graphs $x+4y=24$ and $x+8=y^2.$ This area can be computed in two different ways using integrals

First of all it can be computed as a sum of two integrals

${\displaystyle \underset{a}{\overset{b}{}}}f(x)dx+{\displaystyle \underset{b}{\overset{c}{}}}g(x)dx$

where $a=,b=,c=$ and

$(x+8\}$

$g(x)=$

Alternatively this area can be computed as a single integral

${\displaystyle \underset{}{\overset{}{}}}h(y)dy$

where $=,=$ and

$h(y)=32-4y-y^2$

Either way we find that the area is