Area between the curves $f(x)= x^2-2$ and $g(x)= 3x+2$ My younger sister is a senior in high school and she asked me for help with her calculus homework but I've drawn a blank, can anyone offer any advice for this question?
Find the area between the curves $f(x)= x^2-2$ and $g(x)= 3x+2$.
I vaguely remember this from college but I don't want to tell her the wrong method!  
 A: First, you should graph the two functions just to get an image of them in your head. Next, set them equal to each other to find the interval (You will use integration). Once you set them equal to one another you'll get that the interval is $0 \leq x \leq 4$. The integral sets up as follows:
I'm using S as a symbol to represent integration.
S from $-1$ to $4$ of $g(x)-f(x)dx = S$ from $-1$ to $4$ $$ \int_{-1}^{4}(3x+2)-(x^2-2) dx = ((\frac{-x^3}{3})+((\frac{x^2}{2})+4x)$$
From their you can just plug in. I hope this helps!
A: First find the points where the two functions will intersect.
$$x^2-2 = 3x+2 \implies x^2-3x-4=0\implies x=4,-1$$
Now we take the definite integral of the difference of the functions.
$$\int^{4}_{-1}(3x+2) - (x^2-2) \, dx = \left[\frac{3x^2}{2}-\frac{x^3}{3}+4x\right]^4_{-1}=\boxed{\frac{125}6}$$
A: Figure out the intersections of the two curves and then which one is on top and on bottom. The x values of the intersections will be your bounds and you should take the integral of the top curve minus the bottom curve. 
A: $$A = \int_b^a(g(x)-f(x))dx $$
Where $a$ and $b$ are the points at which the curves intersect where:
$$x^2-2 = 3x+2$$
$$x^2-3x-4=0$$
$$(x-4)(x+1) = 0$$
$$x=4,-1$$
So $$A = \int_b^a(g(x)-f(x))dx = \int_{-1}^4 3x+2-x^2+2 = \int_{-1}^43x-x^2+4 = \left[\frac{3x^2}{2}-\frac{x^3}{3}+4x\right]_{-1}^4 = \boxed{\frac{125}{6}}$$
