Area between four functions I'm not sure about calculating the area between:
$f(x)=x+3$ ,$g(x)=x^2-9$,$k(x)=5$ and $y=0$

My idea , but i'm not sure about the first integral, is:
$$$$
$\int_\sqrt{14}^05-(x-3)dx$ + $\int_0^25-(x+3)dx$
 A: There are some mistakes in your idea: if you look at the picture, the first integral should go from $-\sqrt(14)$ to $-3$ and the second from $-3$ to $2$.
And because in the first integral the bounding functions are $g$ and $k$, you need to integrate $k(x)-g(x)$ instead of $k(x)-f(x)$.
A: You'll want to consider the entire shape as two shapes, divided vertically at the point where your $f(x)$ and $g(x)$ cross. 
Then, you'll compute the area under $k(x)$ but above $g(x)$, and entirely separately, the area under $k(x)$ but above $f(x)$. 
You will then add together these two separate areas to find the total area.
In a general form, this is represented as:
$(\int^a_b{k(x)dx} - \int^a_b{g(x)dx}) + (\int^c_d{f(x)dx} - \int^c_d{g(x)dx})$
Which simplifies to:
$(\int^a_b{k(x)-g(x)dx}) + (\int^c_d{f(x)-g(x)dx})$
Where $a$, $b$, $c$, and $d$ are the intersections of the lines or the division boundary I talked about at the start.
A: it should be 

$$\int _{ -\sqrt { 14 }  }^{ -3 }{ \left( { x }^{ 2 }-9 \right) dx+\int _{ -3 }^{ 2 }{ \left( 5-\left( x+3 \right)  \right) dx }  } $$

